How does one rigorously prove $mathcal{P}({a,b})={emptyset,{a},{b},{a,b}}$












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Clearly, every element of ${emptyset,{a},{b},{a,b}}$ is a subset of ${a,b}$, but I do not see how to rigorously prove the reverse implication. Using the formula for the cardinality of a finite set it can be deduced that $mathcal{P}({a,b})$ has $4$ elements, but surely there must be a more elementary way to prove such an elementary fact?






elementary-set-theory







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  • $begingroup$
    Clarify for me - what exactly is the "reverse implication," here?
    $endgroup$
    – Eevee Trainer
    Dec 7 '18 at 0:45










  • $begingroup$
    I don't think there is much to be shown. A subset of ${a,b}$ clearly has to be one of the sets given.
    $endgroup$
    – Eclipse Sun
    Dec 7 '18 at 0:46










  • $begingroup$
    Just go back to the definition, together with the fact that two sets are equal if and only if they have the same elements. If $X in mathcal{P}({a, b})$, then $X$ is a subset of ${a, b}$. So the elements of $X$ can only consist of $a$ and $b$ and so they must be on the list you wrote.
    $endgroup$
    – T. Bongers
    Dec 7 '18 at 0:46










  • $begingroup$
    @Eevee Trainer How can we be sure that every subset of {a,b} is an element of ${emptyset,{a},{b},{a,b}}$.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:46












  • $begingroup$
    @NAnotapplicable ask yourself is $S subseteq {a,b}$, then $|S| leq 2$. If $|S|=0$, then $S=emptyset$, if $|S|=1$, then either $S={a}$ or $S={b}$.
    $endgroup$
    – Anurag A
    Dec 7 '18 at 0:50
















0












$begingroup$


Clearly, every element of ${emptyset,{a},{b},{a,b}}$ is a subset of ${a,b}$, but I do not see how to rigorously prove the reverse implication. Using the formula for the cardinality of a finite set it can be deduced that $mathcal{P}({a,b})$ has $4$ elements, but surely there must be a more elementary way to prove such an elementary fact?






elementary-set-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Clarify for me - what exactly is the "reverse implication," here?
    $endgroup$
    – Eevee Trainer
    Dec 7 '18 at 0:45










  • $begingroup$
    I don't think there is much to be shown. A subset of ${a,b}$ clearly has to be one of the sets given.
    $endgroup$
    – Eclipse Sun
    Dec 7 '18 at 0:46










  • $begingroup$
    Just go back to the definition, together with the fact that two sets are equal if and only if they have the same elements. If $X in mathcal{P}({a, b})$, then $X$ is a subset of ${a, b}$. So the elements of $X$ can only consist of $a$ and $b$ and so they must be on the list you wrote.
    $endgroup$
    – T. Bongers
    Dec 7 '18 at 0:46










  • $begingroup$
    @Eevee Trainer How can we be sure that every subset of {a,b} is an element of ${emptyset,{a},{b},{a,b}}$.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:46












  • $begingroup$
    @NAnotapplicable ask yourself is $S subseteq {a,b}$, then $|S| leq 2$. If $|S|=0$, then $S=emptyset$, if $|S|=1$, then either $S={a}$ or $S={b}$.
    $endgroup$
    – Anurag A
    Dec 7 '18 at 0:50














0












0








0





$begingroup$


Clearly, every element of ${emptyset,{a},{b},{a,b}}$ is a subset of ${a,b}$, but I do not see how to rigorously prove the reverse implication. Using the formula for the cardinality of a finite set it can be deduced that $mathcal{P}({a,b})$ has $4$ elements, but surely there must be a more elementary way to prove such an elementary fact?






elementary-set-theory







share|cite|improve this question









$endgroup$




Clearly, every element of ${emptyset,{a},{b},{a,b}}$ is a subset of ${a,b}$, but I do not see how to rigorously prove the reverse implication. Using the formula for the cardinality of a finite set it can be deduced that $mathcal{P}({a,b})$ has $4$ elements, but surely there must be a more elementary way to prove such an elementary fact?






elementary-set-theory




elementary-set-theory






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share|cite|improve this question




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asked Dec 7 '18 at 0:44









NAnotapplicableNAnotapplicable

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32












  • $begingroup$
    Clarify for me - what exactly is the "reverse implication," here?
    $endgroup$
    – Eevee Trainer
    Dec 7 '18 at 0:45










  • $begingroup$
    I don't think there is much to be shown. A subset of ${a,b}$ clearly has to be one of the sets given.
    $endgroup$
    – Eclipse Sun
    Dec 7 '18 at 0:46










  • $begingroup$
    Just go back to the definition, together with the fact that two sets are equal if and only if they have the same elements. If $X in mathcal{P}({a, b})$, then $X$ is a subset of ${a, b}$. So the elements of $X$ can only consist of $a$ and $b$ and so they must be on the list you wrote.
    $endgroup$
    – T. Bongers
    Dec 7 '18 at 0:46










  • $begingroup$
    @Eevee Trainer How can we be sure that every subset of {a,b} is an element of ${emptyset,{a},{b},{a,b}}$.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:46












  • $begingroup$
    @NAnotapplicable ask yourself is $S subseteq {a,b}$, then $|S| leq 2$. If $|S|=0$, then $S=emptyset$, if $|S|=1$, then either $S={a}$ or $S={b}$.
    $endgroup$
    – Anurag A
    Dec 7 '18 at 0:50


















  • $begingroup$
    Clarify for me - what exactly is the "reverse implication," here?
    $endgroup$
    – Eevee Trainer
    Dec 7 '18 at 0:45










  • $begingroup$
    I don't think there is much to be shown. A subset of ${a,b}$ clearly has to be one of the sets given.
    $endgroup$
    – Eclipse Sun
    Dec 7 '18 at 0:46










  • $begingroup$
    Just go back to the definition, together with the fact that two sets are equal if and only if they have the same elements. If $X in mathcal{P}({a, b})$, then $X$ is a subset of ${a, b}$. So the elements of $X$ can only consist of $a$ and $b$ and so they must be on the list you wrote.
    $endgroup$
    – T. Bongers
    Dec 7 '18 at 0:46










  • $begingroup$
    @Eevee Trainer How can we be sure that every subset of {a,b} is an element of ${emptyset,{a},{b},{a,b}}$.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:46












  • $begingroup$
    @NAnotapplicable ask yourself is $S subseteq {a,b}$, then $|S| leq 2$. If $|S|=0$, then $S=emptyset$, if $|S|=1$, then either $S={a}$ or $S={b}$.
    $endgroup$
    – Anurag A
    Dec 7 '18 at 0:50
















$begingroup$
Clarify for me - what exactly is the "reverse implication," here?
$endgroup$
– Eevee Trainer
Dec 7 '18 at 0:45




$begingroup$
Clarify for me - what exactly is the "reverse implication," here?
$endgroup$
– Eevee Trainer
Dec 7 '18 at 0:45












$begingroup$
I don't think there is much to be shown. A subset of ${a,b}$ clearly has to be one of the sets given.
$endgroup$
– Eclipse Sun
Dec 7 '18 at 0:46




$begingroup$
I don't think there is much to be shown. A subset of ${a,b}$ clearly has to be one of the sets given.
$endgroup$
– Eclipse Sun
Dec 7 '18 at 0:46












$begingroup$
Just go back to the definition, together with the fact that two sets are equal if and only if they have the same elements. If $X in mathcal{P}({a, b})$, then $X$ is a subset of ${a, b}$. So the elements of $X$ can only consist of $a$ and $b$ and so they must be on the list you wrote.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:46




$begingroup$
Just go back to the definition, together with the fact that two sets are equal if and only if they have the same elements. If $X in mathcal{P}({a, b})$, then $X$ is a subset of ${a, b}$. So the elements of $X$ can only consist of $a$ and $b$ and so they must be on the list you wrote.
$endgroup$
– T. Bongers
Dec 7 '18 at 0:46












$begingroup$
@Eevee Trainer How can we be sure that every subset of {a,b} is an element of ${emptyset,{a},{b},{a,b}}$.
$endgroup$
– NAnotapplicable
Dec 7 '18 at 0:46






$begingroup$
@Eevee Trainer How can we be sure that every subset of {a,b} is an element of ${emptyset,{a},{b},{a,b}}$.
$endgroup$
– NAnotapplicable
Dec 7 '18 at 0:46














$begingroup$
@NAnotapplicable ask yourself is $S subseteq {a,b}$, then $|S| leq 2$. If $|S|=0$, then $S=emptyset$, if $|S|=1$, then either $S={a}$ or $S={b}$.
$endgroup$
– Anurag A
Dec 7 '18 at 0:50




$begingroup$
@NAnotapplicable ask yourself is $S subseteq {a,b}$, then $|S| leq 2$. If $|S|=0$, then $S=emptyset$, if $|S|=1$, then either $S={a}$ or $S={b}$.
$endgroup$
– Anurag A
Dec 7 '18 at 0:50










2 Answers
2






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1












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Suppose $S in mathcal{P}({a,b})$. Then by the law of excluded middle, we have $a in S vee a notin S$; and similarly, we have $b in S vee b notin S$. This splits into four cases: $(ain S wedge bin S) vee (ain S wedge b notin S) vee (anotin S wedge bin S) vee (anotin S wedge bnotin S)$. I'll illustrate one particular case, where $a in S$ and $b notin S$.



Now, for any $x$, suppose $x in S$. Since $S in mathcal{P}({a,b})$, $S subseteq {a,b}$, so this implies $x in {a,b}$, which is equivalent to $x = a vee x = b$. But in the second case, $x = b$, $x in S$, and $b notin S$ leads to a contradiction. Therefore, $x in S rightarrow x = a$; and conversely, $x = a rightarrow x in S$ by substitution with our assumption $a in S$. In conclusion, we have shown $forall x, x in S leftrightarrow x = a leftrightarrow x in { a }$. By the extensionality axiom, this implies that $S = { a }$, so $S in {emptyset,{a},{b},{a,b}}$.



Combining the cases, we will eventually get a proof of $forall S, S in mathcal{P}({a,b}) rightarrow S in {emptyset,{a},{b},{a,b}}$. For the other direction, $S in {emptyset,{a},{b},{a,b}}$ implies $S = emptyset vee S = {a} vee S={b} vee S={a,b}$. I will leave to you to do a proof by cases that each of these implies $S in mathcal{P}({a,b})$.



Therefore, we have $forall S, S in mathcal{P}({a,b}) leftrightarrow S in {emptyset,{a},{b},{a,b}}$, which by another application of the extensionality axiom implies that $mathcal{P}({a,b}) = {emptyset,{a},{b},{a,b}}$.






share|cite|improve this answer









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  • $begingroup$
    Bravo! Exactly what I was looking for.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 1:26










  • $begingroup$
    @NAnotapplicable Out of genuine curiosity, how is this any different than enumeration of the sets in your eyes? During the construction of the cases of consideration, you are forced to explicitly write out every possible subset. The remainder of the proof ends up being even more complicated than the combinatorial argument.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:42












  • $begingroup$
    @RandomMathGuy You are right: the answer I selected is indeed more complicated than yours. I prefer it, however, because it relies directly on the definitions of power set, pair set, etc. rather than introducing the more complicated (as I see it) concept of cardinality, which, although undoubtedly useful, did not seem necessary to prove such a simple assertion. My sense of 'mathematical simplicity' is clearly not the same as yours; both answers have their merits and weaknesses. They're both correct though, (and both appreciated), so which one I prefer is a subjective matter. No hard feelings.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 4:20



















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I think that you are not going to find a more "elementary" way of working with finite powersets precisely because the powerset operation is axiomatic. The fact that such a set exists is given, you don't have to prove it; thus verifying that a given set is a powerset must boil down to enumerating the elements.



Now, I'm no logician, so I could be wrong. If someone out there has better insight then down-vote this into oblivion.






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  • $begingroup$
    But even after enumerating the elements, how can you know that there is not a subset that you missed?
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:54










  • $begingroup$
    @NAnotapplicable If you believe the counting argument for the size of the powerset being $2^{|S|}$, then all you have to do is show that all elements are distinct and that there are the right number of them. If you don't believe the counting argument we will need to go back and address something else that is foundational.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 0:56












  • $begingroup$
    Thank you, I do indeed believe the counting argument and your proof is perfectly valid. Nonetheless, it seems to me that there must be a simpler way to prove this.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:57












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    @NAnotapplicable The issue I'm dancing around is that this is the simple way to prove that a set is a powerset. It is ugly because it is purely computational, but that doesn't make it complicated. I really meant what I wrote; you are dealing with an object that is at the ground floor of ZFC. There is a very good chance that proving a finite set is the powerset of some other set can only be done via enumeration.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 0:59












  • $begingroup$
    Okay then. I admit I'm not fully satisfied but if that really is the simplest way than I guess I never will be... Thanks for your input.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 1:02













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2 Answers
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2 Answers
2






active

oldest

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active

oldest

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active

oldest

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1












$begingroup$

Suppose $S in mathcal{P}({a,b})$. Then by the law of excluded middle, we have $a in S vee a notin S$; and similarly, we have $b in S vee b notin S$. This splits into four cases: $(ain S wedge bin S) vee (ain S wedge b notin S) vee (anotin S wedge bin S) vee (anotin S wedge bnotin S)$. I'll illustrate one particular case, where $a in S$ and $b notin S$.



Now, for any $x$, suppose $x in S$. Since $S in mathcal{P}({a,b})$, $S subseteq {a,b}$, so this implies $x in {a,b}$, which is equivalent to $x = a vee x = b$. But in the second case, $x = b$, $x in S$, and $b notin S$ leads to a contradiction. Therefore, $x in S rightarrow x = a$; and conversely, $x = a rightarrow x in S$ by substitution with our assumption $a in S$. In conclusion, we have shown $forall x, x in S leftrightarrow x = a leftrightarrow x in { a }$. By the extensionality axiom, this implies that $S = { a }$, so $S in {emptyset,{a},{b},{a,b}}$.



Combining the cases, we will eventually get a proof of $forall S, S in mathcal{P}({a,b}) rightarrow S in {emptyset,{a},{b},{a,b}}$. For the other direction, $S in {emptyset,{a},{b},{a,b}}$ implies $S = emptyset vee S = {a} vee S={b} vee S={a,b}$. I will leave to you to do a proof by cases that each of these implies $S in mathcal{P}({a,b})$.



Therefore, we have $forall S, S in mathcal{P}({a,b}) leftrightarrow S in {emptyset,{a},{b},{a,b}}$, which by another application of the extensionality axiom implies that $mathcal{P}({a,b}) = {emptyset,{a},{b},{a,b}}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Bravo! Exactly what I was looking for.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 1:26










  • $begingroup$
    @NAnotapplicable Out of genuine curiosity, how is this any different than enumeration of the sets in your eyes? During the construction of the cases of consideration, you are forced to explicitly write out every possible subset. The remainder of the proof ends up being even more complicated than the combinatorial argument.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:42












  • $begingroup$
    @RandomMathGuy You are right: the answer I selected is indeed more complicated than yours. I prefer it, however, because it relies directly on the definitions of power set, pair set, etc. rather than introducing the more complicated (as I see it) concept of cardinality, which, although undoubtedly useful, did not seem necessary to prove such a simple assertion. My sense of 'mathematical simplicity' is clearly not the same as yours; both answers have their merits and weaknesses. They're both correct though, (and both appreciated), so which one I prefer is a subjective matter. No hard feelings.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 4:20
















1












$begingroup$

Suppose $S in mathcal{P}({a,b})$. Then by the law of excluded middle, we have $a in S vee a notin S$; and similarly, we have $b in S vee b notin S$. This splits into four cases: $(ain S wedge bin S) vee (ain S wedge b notin S) vee (anotin S wedge bin S) vee (anotin S wedge bnotin S)$. I'll illustrate one particular case, where $a in S$ and $b notin S$.



Now, for any $x$, suppose $x in S$. Since $S in mathcal{P}({a,b})$, $S subseteq {a,b}$, so this implies $x in {a,b}$, which is equivalent to $x = a vee x = b$. But in the second case, $x = b$, $x in S$, and $b notin S$ leads to a contradiction. Therefore, $x in S rightarrow x = a$; and conversely, $x = a rightarrow x in S$ by substitution with our assumption $a in S$. In conclusion, we have shown $forall x, x in S leftrightarrow x = a leftrightarrow x in { a }$. By the extensionality axiom, this implies that $S = { a }$, so $S in {emptyset,{a},{b},{a,b}}$.



Combining the cases, we will eventually get a proof of $forall S, S in mathcal{P}({a,b}) rightarrow S in {emptyset,{a},{b},{a,b}}$. For the other direction, $S in {emptyset,{a},{b},{a,b}}$ implies $S = emptyset vee S = {a} vee S={b} vee S={a,b}$. I will leave to you to do a proof by cases that each of these implies $S in mathcal{P}({a,b})$.



Therefore, we have $forall S, S in mathcal{P}({a,b}) leftrightarrow S in {emptyset,{a},{b},{a,b}}$, which by another application of the extensionality axiom implies that $mathcal{P}({a,b}) = {emptyset,{a},{b},{a,b}}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Bravo! Exactly what I was looking for.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 1:26










  • $begingroup$
    @NAnotapplicable Out of genuine curiosity, how is this any different than enumeration of the sets in your eyes? During the construction of the cases of consideration, you are forced to explicitly write out every possible subset. The remainder of the proof ends up being even more complicated than the combinatorial argument.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:42












  • $begingroup$
    @RandomMathGuy You are right: the answer I selected is indeed more complicated than yours. I prefer it, however, because it relies directly on the definitions of power set, pair set, etc. rather than introducing the more complicated (as I see it) concept of cardinality, which, although undoubtedly useful, did not seem necessary to prove such a simple assertion. My sense of 'mathematical simplicity' is clearly not the same as yours; both answers have their merits and weaknesses. They're both correct though, (and both appreciated), so which one I prefer is a subjective matter. No hard feelings.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 4:20














1












1








1





$begingroup$

Suppose $S in mathcal{P}({a,b})$. Then by the law of excluded middle, we have $a in S vee a notin S$; and similarly, we have $b in S vee b notin S$. This splits into four cases: $(ain S wedge bin S) vee (ain S wedge b notin S) vee (anotin S wedge bin S) vee (anotin S wedge bnotin S)$. I'll illustrate one particular case, where $a in S$ and $b notin S$.



Now, for any $x$, suppose $x in S$. Since $S in mathcal{P}({a,b})$, $S subseteq {a,b}$, so this implies $x in {a,b}$, which is equivalent to $x = a vee x = b$. But in the second case, $x = b$, $x in S$, and $b notin S$ leads to a contradiction. Therefore, $x in S rightarrow x = a$; and conversely, $x = a rightarrow x in S$ by substitution with our assumption $a in S$. In conclusion, we have shown $forall x, x in S leftrightarrow x = a leftrightarrow x in { a }$. By the extensionality axiom, this implies that $S = { a }$, so $S in {emptyset,{a},{b},{a,b}}$.



Combining the cases, we will eventually get a proof of $forall S, S in mathcal{P}({a,b}) rightarrow S in {emptyset,{a},{b},{a,b}}$. For the other direction, $S in {emptyset,{a},{b},{a,b}}$ implies $S = emptyset vee S = {a} vee S={b} vee S={a,b}$. I will leave to you to do a proof by cases that each of these implies $S in mathcal{P}({a,b})$.



Therefore, we have $forall S, S in mathcal{P}({a,b}) leftrightarrow S in {emptyset,{a},{b},{a,b}}$, which by another application of the extensionality axiom implies that $mathcal{P}({a,b}) = {emptyset,{a},{b},{a,b}}$.






share|cite|improve this answer









$endgroup$



Suppose $S in mathcal{P}({a,b})$. Then by the law of excluded middle, we have $a in S vee a notin S$; and similarly, we have $b in S vee b notin S$. This splits into four cases: $(ain S wedge bin S) vee (ain S wedge b notin S) vee (anotin S wedge bin S) vee (anotin S wedge bnotin S)$. I'll illustrate one particular case, where $a in S$ and $b notin S$.



Now, for any $x$, suppose $x in S$. Since $S in mathcal{P}({a,b})$, $S subseteq {a,b}$, so this implies $x in {a,b}$, which is equivalent to $x = a vee x = b$. But in the second case, $x = b$, $x in S$, and $b notin S$ leads to a contradiction. Therefore, $x in S rightarrow x = a$; and conversely, $x = a rightarrow x in S$ by substitution with our assumption $a in S$. In conclusion, we have shown $forall x, x in S leftrightarrow x = a leftrightarrow x in { a }$. By the extensionality axiom, this implies that $S = { a }$, so $S in {emptyset,{a},{b},{a,b}}$.



Combining the cases, we will eventually get a proof of $forall S, S in mathcal{P}({a,b}) rightarrow S in {emptyset,{a},{b},{a,b}}$. For the other direction, $S in {emptyset,{a},{b},{a,b}}$ implies $S = emptyset vee S = {a} vee S={b} vee S={a,b}$. I will leave to you to do a proof by cases that each of these implies $S in mathcal{P}({a,b})$.



Therefore, we have $forall S, S in mathcal{P}({a,b}) leftrightarrow S in {emptyset,{a},{b},{a,b}}$, which by another application of the extensionality axiom implies that $mathcal{P}({a,b}) = {emptyset,{a},{b},{a,b}}$.







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share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 1:17









Daniel ScheplerDaniel Schepler

8,7291620




8,7291620












  • $begingroup$
    Bravo! Exactly what I was looking for.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 1:26










  • $begingroup$
    @NAnotapplicable Out of genuine curiosity, how is this any different than enumeration of the sets in your eyes? During the construction of the cases of consideration, you are forced to explicitly write out every possible subset. The remainder of the proof ends up being even more complicated than the combinatorial argument.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:42












  • $begingroup$
    @RandomMathGuy You are right: the answer I selected is indeed more complicated than yours. I prefer it, however, because it relies directly on the definitions of power set, pair set, etc. rather than introducing the more complicated (as I see it) concept of cardinality, which, although undoubtedly useful, did not seem necessary to prove such a simple assertion. My sense of 'mathematical simplicity' is clearly not the same as yours; both answers have their merits and weaknesses. They're both correct though, (and both appreciated), so which one I prefer is a subjective matter. No hard feelings.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 4:20


















  • $begingroup$
    Bravo! Exactly what I was looking for.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 1:26










  • $begingroup$
    @NAnotapplicable Out of genuine curiosity, how is this any different than enumeration of the sets in your eyes? During the construction of the cases of consideration, you are forced to explicitly write out every possible subset. The remainder of the proof ends up being even more complicated than the combinatorial argument.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 2:42












  • $begingroup$
    @RandomMathGuy You are right: the answer I selected is indeed more complicated than yours. I prefer it, however, because it relies directly on the definitions of power set, pair set, etc. rather than introducing the more complicated (as I see it) concept of cardinality, which, although undoubtedly useful, did not seem necessary to prove such a simple assertion. My sense of 'mathematical simplicity' is clearly not the same as yours; both answers have their merits and weaknesses. They're both correct though, (and both appreciated), so which one I prefer is a subjective matter. No hard feelings.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 4:20
















$begingroup$
Bravo! Exactly what I was looking for.
$endgroup$
– NAnotapplicable
Dec 7 '18 at 1:26




$begingroup$
Bravo! Exactly what I was looking for.
$endgroup$
– NAnotapplicable
Dec 7 '18 at 1:26












$begingroup$
@NAnotapplicable Out of genuine curiosity, how is this any different than enumeration of the sets in your eyes? During the construction of the cases of consideration, you are forced to explicitly write out every possible subset. The remainder of the proof ends up being even more complicated than the combinatorial argument.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:42






$begingroup$
@NAnotapplicable Out of genuine curiosity, how is this any different than enumeration of the sets in your eyes? During the construction of the cases of consideration, you are forced to explicitly write out every possible subset. The remainder of the proof ends up being even more complicated than the combinatorial argument.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 2:42














$begingroup$
@RandomMathGuy You are right: the answer I selected is indeed more complicated than yours. I prefer it, however, because it relies directly on the definitions of power set, pair set, etc. rather than introducing the more complicated (as I see it) concept of cardinality, which, although undoubtedly useful, did not seem necessary to prove such a simple assertion. My sense of 'mathematical simplicity' is clearly not the same as yours; both answers have their merits and weaknesses. They're both correct though, (and both appreciated), so which one I prefer is a subjective matter. No hard feelings.
$endgroup$
– NAnotapplicable
Dec 7 '18 at 4:20




$begingroup$
@RandomMathGuy You are right: the answer I selected is indeed more complicated than yours. I prefer it, however, because it relies directly on the definitions of power set, pair set, etc. rather than introducing the more complicated (as I see it) concept of cardinality, which, although undoubtedly useful, did not seem necessary to prove such a simple assertion. My sense of 'mathematical simplicity' is clearly not the same as yours; both answers have their merits and weaknesses. They're both correct though, (and both appreciated), so which one I prefer is a subjective matter. No hard feelings.
$endgroup$
– NAnotapplicable
Dec 7 '18 at 4:20











1












$begingroup$

I think that you are not going to find a more "elementary" way of working with finite powersets precisely because the powerset operation is axiomatic. The fact that such a set exists is given, you don't have to prove it; thus verifying that a given set is a powerset must boil down to enumerating the elements.



Now, I'm no logician, so I could be wrong. If someone out there has better insight then down-vote this into oblivion.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But even after enumerating the elements, how can you know that there is not a subset that you missed?
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:54










  • $begingroup$
    @NAnotapplicable If you believe the counting argument for the size of the powerset being $2^{|S|}$, then all you have to do is show that all elements are distinct and that there are the right number of them. If you don't believe the counting argument we will need to go back and address something else that is foundational.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 0:56












  • $begingroup$
    Thank you, I do indeed believe the counting argument and your proof is perfectly valid. Nonetheless, it seems to me that there must be a simpler way to prove this.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:57












  • $begingroup$
    @NAnotapplicable The issue I'm dancing around is that this is the simple way to prove that a set is a powerset. It is ugly because it is purely computational, but that doesn't make it complicated. I really meant what I wrote; you are dealing with an object that is at the ground floor of ZFC. There is a very good chance that proving a finite set is the powerset of some other set can only be done via enumeration.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 0:59












  • $begingroup$
    Okay then. I admit I'm not fully satisfied but if that really is the simplest way than I guess I never will be... Thanks for your input.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 1:02


















1












$begingroup$

I think that you are not going to find a more "elementary" way of working with finite powersets precisely because the powerset operation is axiomatic. The fact that such a set exists is given, you don't have to prove it; thus verifying that a given set is a powerset must boil down to enumerating the elements.



Now, I'm no logician, so I could be wrong. If someone out there has better insight then down-vote this into oblivion.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But even after enumerating the elements, how can you know that there is not a subset that you missed?
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:54










  • $begingroup$
    @NAnotapplicable If you believe the counting argument for the size of the powerset being $2^{|S|}$, then all you have to do is show that all elements are distinct and that there are the right number of them. If you don't believe the counting argument we will need to go back and address something else that is foundational.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 0:56












  • $begingroup$
    Thank you, I do indeed believe the counting argument and your proof is perfectly valid. Nonetheless, it seems to me that there must be a simpler way to prove this.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:57












  • $begingroup$
    @NAnotapplicable The issue I'm dancing around is that this is the simple way to prove that a set is a powerset. It is ugly because it is purely computational, but that doesn't make it complicated. I really meant what I wrote; you are dealing with an object that is at the ground floor of ZFC. There is a very good chance that proving a finite set is the powerset of some other set can only be done via enumeration.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 0:59












  • $begingroup$
    Okay then. I admit I'm not fully satisfied but if that really is the simplest way than I guess I never will be... Thanks for your input.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 1:02
















1












1








1





$begingroup$

I think that you are not going to find a more "elementary" way of working with finite powersets precisely because the powerset operation is axiomatic. The fact that such a set exists is given, you don't have to prove it; thus verifying that a given set is a powerset must boil down to enumerating the elements.



Now, I'm no logician, so I could be wrong. If someone out there has better insight then down-vote this into oblivion.






share|cite|improve this answer









$endgroup$



I think that you are not going to find a more "elementary" way of working with finite powersets precisely because the powerset operation is axiomatic. The fact that such a set exists is given, you don't have to prove it; thus verifying that a given set is a powerset must boil down to enumerating the elements.



Now, I'm no logician, so I could be wrong. If someone out there has better insight then down-vote this into oblivion.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 0:51









RandomMathGuyRandomMathGuy

1192




1192












  • $begingroup$
    But even after enumerating the elements, how can you know that there is not a subset that you missed?
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:54










  • $begingroup$
    @NAnotapplicable If you believe the counting argument for the size of the powerset being $2^{|S|}$, then all you have to do is show that all elements are distinct and that there are the right number of them. If you don't believe the counting argument we will need to go back and address something else that is foundational.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 0:56












  • $begingroup$
    Thank you, I do indeed believe the counting argument and your proof is perfectly valid. Nonetheless, it seems to me that there must be a simpler way to prove this.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:57












  • $begingroup$
    @NAnotapplicable The issue I'm dancing around is that this is the simple way to prove that a set is a powerset. It is ugly because it is purely computational, but that doesn't make it complicated. I really meant what I wrote; you are dealing with an object that is at the ground floor of ZFC. There is a very good chance that proving a finite set is the powerset of some other set can only be done via enumeration.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 0:59












  • $begingroup$
    Okay then. I admit I'm not fully satisfied but if that really is the simplest way than I guess I never will be... Thanks for your input.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 1:02




















  • $begingroup$
    But even after enumerating the elements, how can you know that there is not a subset that you missed?
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:54










  • $begingroup$
    @NAnotapplicable If you believe the counting argument for the size of the powerset being $2^{|S|}$, then all you have to do is show that all elements are distinct and that there are the right number of them. If you don't believe the counting argument we will need to go back and address something else that is foundational.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 0:56












  • $begingroup$
    Thank you, I do indeed believe the counting argument and your proof is perfectly valid. Nonetheless, it seems to me that there must be a simpler way to prove this.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 0:57












  • $begingroup$
    @NAnotapplicable The issue I'm dancing around is that this is the simple way to prove that a set is a powerset. It is ugly because it is purely computational, but that doesn't make it complicated. I really meant what I wrote; you are dealing with an object that is at the ground floor of ZFC. There is a very good chance that proving a finite set is the powerset of some other set can only be done via enumeration.
    $endgroup$
    – RandomMathGuy
    Dec 7 '18 at 0:59












  • $begingroup$
    Okay then. I admit I'm not fully satisfied but if that really is the simplest way than I guess I never will be... Thanks for your input.
    $endgroup$
    – NAnotapplicable
    Dec 7 '18 at 1:02


















$begingroup$
But even after enumerating the elements, how can you know that there is not a subset that you missed?
$endgroup$
– NAnotapplicable
Dec 7 '18 at 0:54




$begingroup$
But even after enumerating the elements, how can you know that there is not a subset that you missed?
$endgroup$
– NAnotapplicable
Dec 7 '18 at 0:54












$begingroup$
@NAnotapplicable If you believe the counting argument for the size of the powerset being $2^{|S|}$, then all you have to do is show that all elements are distinct and that there are the right number of them. If you don't believe the counting argument we will need to go back and address something else that is foundational.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 0:56






$begingroup$
@NAnotapplicable If you believe the counting argument for the size of the powerset being $2^{|S|}$, then all you have to do is show that all elements are distinct and that there are the right number of them. If you don't believe the counting argument we will need to go back and address something else that is foundational.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 0:56














$begingroup$
Thank you, I do indeed believe the counting argument and your proof is perfectly valid. Nonetheless, it seems to me that there must be a simpler way to prove this.
$endgroup$
– NAnotapplicable
Dec 7 '18 at 0:57






$begingroup$
Thank you, I do indeed believe the counting argument and your proof is perfectly valid. Nonetheless, it seems to me that there must be a simpler way to prove this.
$endgroup$
– NAnotapplicable
Dec 7 '18 at 0:57














$begingroup$
@NAnotapplicable The issue I'm dancing around is that this is the simple way to prove that a set is a powerset. It is ugly because it is purely computational, but that doesn't make it complicated. I really meant what I wrote; you are dealing with an object that is at the ground floor of ZFC. There is a very good chance that proving a finite set is the powerset of some other set can only be done via enumeration.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 0:59






$begingroup$
@NAnotapplicable The issue I'm dancing around is that this is the simple way to prove that a set is a powerset. It is ugly because it is purely computational, but that doesn't make it complicated. I really meant what I wrote; you are dealing with an object that is at the ground floor of ZFC. There is a very good chance that proving a finite set is the powerset of some other set can only be done via enumeration.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 0:59














$begingroup$
Okay then. I admit I'm not fully satisfied but if that really is the simplest way than I guess I never will be... Thanks for your input.
$endgroup$
– NAnotapplicable
Dec 7 '18 at 1:02






$begingroup$
Okay then. I admit I'm not fully satisfied but if that really is the simplest way than I guess I never will be... Thanks for your input.
$endgroup$
– NAnotapplicable
Dec 7 '18 at 1:02




















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