If $Ksubseteq mathbb{R}$ is compact, then it is closed.












1












$begingroup$


We say that $Ksubseteqmathbb{R}$ is compact if for every open cover there exists a finite subcover.



We say that $Ksubseteq mathbb{R}$ is bounded if there exists $M>0$ such that for all $zin K,$ $|z|leq M.$





Question: Show that if $K$ is compact, then $K$ is bounded.




My attempt:



We can assume that $K$ is nonempty.
Otherwise, let $M=1$ and we are done.



Let $xin K.$
Consider the collection
$$ mathcal{O} = { (x-r,x+r): r>0 }.
$$
Clearly $mathcal{O}$ is an open cover for $K.$
By compactness of $K$, there exists a natural number $N$ such that
$$K subseteq bigcup_{k=1}^N (x-r_k,x+r_k).$$
where $r_1,...,r_Nin mathbb{R}_+.$
Let
$$M =|x| + max_{1leq kleq N} r_k.$$
Clearly $M>0.$
Note that for every $zin K,$ we have
$$|z-x| < r_k$$ for some $1leq kleq N.$
By reverse triangle inequality, we have
$$|z| - |x| leq |z-x| < r_k leq M-|x|.$$
It follows that $|z|leq M$ for all $zin K.$
Hence, $K$ is bounded.





My question is that can we take such $M$ in the proof above?
Is it valid?







real-analysis general-topology proof-verification







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Yes that's a good proof and there is no problem with the choice of $M$.
    $endgroup$
    – Yanko
    Dec 15 '18 at 14:54
















1












$begingroup$


We say that $Ksubseteqmathbb{R}$ is compact if for every open cover there exists a finite subcover.



We say that $Ksubseteq mathbb{R}$ is bounded if there exists $M>0$ such that for all $zin K,$ $|z|leq M.$





Question: Show that if $K$ is compact, then $K$ is bounded.




My attempt:



We can assume that $K$ is nonempty.
Otherwise, let $M=1$ and we are done.



Let $xin K.$
Consider the collection
$$ mathcal{O} = { (x-r,x+r): r>0 }.
$$
Clearly $mathcal{O}$ is an open cover for $K.$
By compactness of $K$, there exists a natural number $N$ such that
$$K subseteq bigcup_{k=1}^N (x-r_k,x+r_k).$$
where $r_1,...,r_Nin mathbb{R}_+.$
Let
$$M =|x| + max_{1leq kleq N} r_k.$$
Clearly $M>0.$
Note that for every $zin K,$ we have
$$|z-x| < r_k$$ for some $1leq kleq N.$
By reverse triangle inequality, we have
$$|z| - |x| leq |z-x| < r_k leq M-|x|.$$
It follows that $|z|leq M$ for all $zin K.$
Hence, $K$ is bounded.





My question is that can we take such $M$ in the proof above?
Is it valid?







real-analysis general-topology proof-verification







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Yes that's a good proof and there is no problem with the choice of $M$.
    $endgroup$
    – Yanko
    Dec 15 '18 at 14:54














1












1








1





$begingroup$


We say that $Ksubseteqmathbb{R}$ is compact if for every open cover there exists a finite subcover.



We say that $Ksubseteq mathbb{R}$ is bounded if there exists $M>0$ such that for all $zin K,$ $|z|leq M.$





Question: Show that if $K$ is compact, then $K$ is bounded.




My attempt:



We can assume that $K$ is nonempty.
Otherwise, let $M=1$ and we are done.



Let $xin K.$
Consider the collection
$$ mathcal{O} = { (x-r,x+r): r>0 }.
$$
Clearly $mathcal{O}$ is an open cover for $K.$
By compactness of $K$, there exists a natural number $N$ such that
$$K subseteq bigcup_{k=1}^N (x-r_k,x+r_k).$$
where $r_1,...,r_Nin mathbb{R}_+.$
Let
$$M =|x| + max_{1leq kleq N} r_k.$$
Clearly $M>0.$
Note that for every $zin K,$ we have
$$|z-x| < r_k$$ for some $1leq kleq N.$
By reverse triangle inequality, we have
$$|z| - |x| leq |z-x| < r_k leq M-|x|.$$
It follows that $|z|leq M$ for all $zin K.$
Hence, $K$ is bounded.





My question is that can we take such $M$ in the proof above?
Is it valid?







real-analysis general-topology proof-verification







share|cite|improve this question









$endgroup$




We say that $Ksubseteqmathbb{R}$ is compact if for every open cover there exists a finite subcover.



We say that $Ksubseteq mathbb{R}$ is bounded if there exists $M>0$ such that for all $zin K,$ $|z|leq M.$





Question: Show that if $K$ is compact, then $K$ is bounded.




My attempt:



We can assume that $K$ is nonempty.
Otherwise, let $M=1$ and we are done.



Let $xin K.$
Consider the collection
$$ mathcal{O} = { (x-r,x+r): r>0 }.
$$
Clearly $mathcal{O}$ is an open cover for $K.$
By compactness of $K$, there exists a natural number $N$ such that
$$K subseteq bigcup_{k=1}^N (x-r_k,x+r_k).$$
where $r_1,...,r_Nin mathbb{R}_+.$
Let
$$M =|x| + max_{1leq kleq N} r_k.$$
Clearly $M>0.$
Note that for every $zin K,$ we have
$$|z-x| < r_k$$ for some $1leq kleq N.$
By reverse triangle inequality, we have
$$|z| - |x| leq |z-x| < r_k leq M-|x|.$$
It follows that $|z|leq M$ for all $zin K.$
Hence, $K$ is bounded.





My question is that can we take such $M$ in the proof above?
Is it valid?







real-analysis general-topology proof-verification




real-analysis general-topology proof-verification






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 '18 at 14:46









IdonknowIdonknow

2,504850114




2,504850114








  • 3




    $begingroup$
    Yes that's a good proof and there is no problem with the choice of $M$.
    $endgroup$
    – Yanko
    Dec 15 '18 at 14:54














  • 3




    $begingroup$
    Yes that's a good proof and there is no problem with the choice of $M$.
    $endgroup$
    – Yanko
    Dec 15 '18 at 14:54








3




3




$begingroup$
Yes that's a good proof and there is no problem with the choice of $M$.
$endgroup$
– Yanko
Dec 15 '18 at 14:54




$begingroup$
Yes that's a good proof and there is no problem with the choice of $M$.
$endgroup$
– Yanko
Dec 15 '18 at 14:54










1 Answer
1






active

oldest

votes


















2












$begingroup$

I think it's worthwhile adding that although there is no issue in the proof, an easier method is to take the open cover $mathcal{O} = {(-R, R) mid R > 0}$.



The argument then simplifies to the following: $K$ is compact, so there is a finite subcover $${(-R_i, R_i) mid i = 1, cdots, n}$$



If $overline{R} = max R_i$ then $K subset B_{overline{R}}(0) = (-overline{R}, overline{R})$, i.e. each $x in K$ satisfies $lvert x rvert leq overline{R}$. Note it is true in general that a compact subset of a metric space is bounded (and totally bounded).






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    We know that $R_n$ is the biggest among all $R_i$. So the notation $overline{R}$ is not needed.
    $endgroup$
    – Idonknow
    Dec 16 '18 at 13:48










  • $begingroup$
    Thanks - I'll leave the answer as it is since I did not explicitly state the $R_i$ were ordered by magnitude, but I agree its neater had I written $0 < R_1 < R_2 < cdots < R_n$.
    $endgroup$
    – Riley
    Dec 16 '18 at 13:56













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

I think it's worthwhile adding that although there is no issue in the proof, an easier method is to take the open cover $mathcal{O} = {(-R, R) mid R > 0}$.



The argument then simplifies to the following: $K$ is compact, so there is a finite subcover $${(-R_i, R_i) mid i = 1, cdots, n}$$



If $overline{R} = max R_i$ then $K subset B_{overline{R}}(0) = (-overline{R}, overline{R})$, i.e. each $x in K$ satisfies $lvert x rvert leq overline{R}$. Note it is true in general that a compact subset of a metric space is bounded (and totally bounded).






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    We know that $R_n$ is the biggest among all $R_i$. So the notation $overline{R}$ is not needed.
    $endgroup$
    – Idonknow
    Dec 16 '18 at 13:48










  • $begingroup$
    Thanks - I'll leave the answer as it is since I did not explicitly state the $R_i$ were ordered by magnitude, but I agree its neater had I written $0 < R_1 < R_2 < cdots < R_n$.
    $endgroup$
    – Riley
    Dec 16 '18 at 13:56


















2












$begingroup$

I think it's worthwhile adding that although there is no issue in the proof, an easier method is to take the open cover $mathcal{O} = {(-R, R) mid R > 0}$.



The argument then simplifies to the following: $K$ is compact, so there is a finite subcover $${(-R_i, R_i) mid i = 1, cdots, n}$$



If $overline{R} = max R_i$ then $K subset B_{overline{R}}(0) = (-overline{R}, overline{R})$, i.e. each $x in K$ satisfies $lvert x rvert leq overline{R}$. Note it is true in general that a compact subset of a metric space is bounded (and totally bounded).






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    We know that $R_n$ is the biggest among all $R_i$. So the notation $overline{R}$ is not needed.
    $endgroup$
    – Idonknow
    Dec 16 '18 at 13:48










  • $begingroup$
    Thanks - I'll leave the answer as it is since I did not explicitly state the $R_i$ were ordered by magnitude, but I agree its neater had I written $0 < R_1 < R_2 < cdots < R_n$.
    $endgroup$
    – Riley
    Dec 16 '18 at 13:56
















2












2








2





$begingroup$

I think it's worthwhile adding that although there is no issue in the proof, an easier method is to take the open cover $mathcal{O} = {(-R, R) mid R > 0}$.



The argument then simplifies to the following: $K$ is compact, so there is a finite subcover $${(-R_i, R_i) mid i = 1, cdots, n}$$



If $overline{R} = max R_i$ then $K subset B_{overline{R}}(0) = (-overline{R}, overline{R})$, i.e. each $x in K$ satisfies $lvert x rvert leq overline{R}$. Note it is true in general that a compact subset of a metric space is bounded (and totally bounded).






share|cite|improve this answer









$endgroup$



I think it's worthwhile adding that although there is no issue in the proof, an easier method is to take the open cover $mathcal{O} = {(-R, R) mid R > 0}$.



The argument then simplifies to the following: $K$ is compact, so there is a finite subcover $${(-R_i, R_i) mid i = 1, cdots, n}$$



If $overline{R} = max R_i$ then $K subset B_{overline{R}}(0) = (-overline{R}, overline{R})$, i.e. each $x in K$ satisfies $lvert x rvert leq overline{R}$. Note it is true in general that a compact subset of a metric space is bounded (and totally bounded).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 13:28









RileyRiley

1825




1825








  • 1




    $begingroup$
    We know that $R_n$ is the biggest among all $R_i$. So the notation $overline{R}$ is not needed.
    $endgroup$
    – Idonknow
    Dec 16 '18 at 13:48










  • $begingroup$
    Thanks - I'll leave the answer as it is since I did not explicitly state the $R_i$ were ordered by magnitude, but I agree its neater had I written $0 < R_1 < R_2 < cdots < R_n$.
    $endgroup$
    – Riley
    Dec 16 '18 at 13:56
















  • 1




    $begingroup$
    We know that $R_n$ is the biggest among all $R_i$. So the notation $overline{R}$ is not needed.
    $endgroup$
    – Idonknow
    Dec 16 '18 at 13:48










  • $begingroup$
    Thanks - I'll leave the answer as it is since I did not explicitly state the $R_i$ were ordered by magnitude, but I agree its neater had I written $0 < R_1 < R_2 < cdots < R_n$.
    $endgroup$
    – Riley
    Dec 16 '18 at 13:56










1




1




$begingroup$
We know that $R_n$ is the biggest among all $R_i$. So the notation $overline{R}$ is not needed.
$endgroup$
– Idonknow
Dec 16 '18 at 13:48




$begingroup$
We know that $R_n$ is the biggest among all $R_i$. So the notation $overline{R}$ is not needed.
$endgroup$
– Idonknow
Dec 16 '18 at 13:48












$begingroup$
Thanks - I'll leave the answer as it is since I did not explicitly state the $R_i$ were ordered by magnitude, but I agree its neater had I written $0 < R_1 < R_2 < cdots < R_n$.
$endgroup$
– Riley
Dec 16 '18 at 13:56






$begingroup$
Thanks - I'll leave the answer as it is since I did not explicitly state the $R_i$ were ordered by magnitude, but I agree its neater had I written $0 < R_1 < R_2 < cdots < R_n$.
$endgroup$
– Riley
Dec 16 '18 at 13:56




















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