Proving finite product of discrete topology is discrete
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Exercise: If $(X,tau_1)$,$(X,tau_2)$....$(Xtau_n)$ are discrete spaces. Prove that the product space $(X,tau_1)times(X,tau_2)....times(X,tau_n)$ is also a discrete space.
I had a very straightforward idea that I do not know if it is right.
${a_1,a_2,...a_n,:a_iintau_i,i=1,2...n}$ forms a basis for the topology $tau_n$ that can generate any element then the topology $tau_n$ must be discrete.
Question:
Is my proof right? What are other alternative proofs?
Thanks in advance!
general-topology
asked Dec 18 '18 at 12:02
Pedro GomesPedro Gomes
1,9252721
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add a comment |
$begingroup$
Exercise: If $(X,tau_1)$,$(X,tau_2)$....$(Xtau_n)$ are discrete spaces. Prove that the product space $(X,tau_1)times(X,tau_2)....times(X,tau_n)$ is also a discrete space.
I had a very straightforward idea that I do not know if it is right.
${a_1,a_2,...a_n,:a_iintau_i,i=1,2...n}$ forms a basis for the topology $tau_n$ that can generate any element then the topology $tau_n$ must be discrete.
Question:
Is my proof right? What are other alternative proofs?
Thanks in advance!
general-topology
asked Dec 18 '18 at 12:02
Pedro GomesPedro Gomes
1,9252721
$endgroup$
add a comment |
$begingroup$
Exercise: If $(X,tau_1)$,$(X,tau_2)$....$(Xtau_n)$ are discrete spaces. Prove that the product space $(X,tau_1)times(X,tau_2)....times(X,tau_n)$ is also a discrete space.
I had a very straightforward idea that I do not know if it is right.
${a_1,a_2,...a_n,:a_iintau_i,i=1,2...n}$ forms a basis for the topology $tau_n$ that can generate any element then the topology $tau_n$ must be discrete.
Question:
Is my proof right? What are other alternative proofs?
Thanks in advance!
general-topology
asked Dec 18 '18 at 12:02
Pedro GomesPedro Gomes
1,9252721
$endgroup$
Exercise: If $(X,tau_1)$,$(X,tau_2)$....$(Xtau_n)$ are discrete spaces. Prove that the product space $(X,tau_1)times(X,tau_2)....times(X,tau_n)$ is also a discrete space.
I had a very straightforward idea that I do not know if it is right.
${a_1,a_2,...a_n,:a_iintau_i,i=1,2...n}$ forms a basis for the topology $tau_n$ that can generate any element then the topology $tau_n$ must be discrete.
Question:
Is my proof right? What are other alternative proofs?
Thanks in advance!
general-topology
general-topology
asked Dec 18 '18 at 12:02
Pedro GomesPedro Gomes
1,9252721
asked Dec 18 '18 at 12:02
Pedro GomesPedro Gomes
1,9252721
asked Dec 18 '18 at 12:02
Pedro GomesPedro Gomes
1,9252721
asked Dec 18 '18 at 12:02
Pedro GomesPedro Gomes
1,9252721
asked Dec 18 '18 at 12:02
Pedro GomesPedro Gomes
1,9252721
1,9252721
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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Your proof makes no sense, because the set that you defined makes no sense.
You can prove it as follows: is $p_1in X_1$, $p_2in X_2$, …, $p_nin X_n$, then each set ${p_i}$ is open and therefore, $bigl{(p_1,p_2,ldots,p_n)bigr}$ is an open subset of $prod_{i=1}^nX_i$, since it is equal to ${p_1}times{p_2}timescdotstimes{p_n}$. Since each singleton is open, your topology is discrete.
edited Dec 18 '18 at 12:20
Henno Brandsma
112k348121
answered Dec 18 '18 at 12:10
José Carlos SantosJosé Carlos Santos
168k22132236
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your proof makes no sense, because the set that you defined makes no sense.
You can prove it as follows: is $p_1in X_1$, $p_2in X_2$, …, $p_nin X_n$, then each set ${p_i}$ is open and therefore, $bigl{(p_1,p_2,ldots,p_n)bigr}$ is an open subset of $prod_{i=1}^nX_i$, since it is equal to ${p_1}times{p_2}timescdotstimes{p_n}$. Since each singleton is open, your topology is discrete.
edited Dec 18 '18 at 12:20
Henno Brandsma
112k348121
answered Dec 18 '18 at 12:10
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
Your proof makes no sense, because the set that you defined makes no sense.
You can prove it as follows: is $p_1in X_1$, $p_2in X_2$, …, $p_nin X_n$, then each set ${p_i}$ is open and therefore, $bigl{(p_1,p_2,ldots,p_n)bigr}$ is an open subset of $prod_{i=1}^nX_i$, since it is equal to ${p_1}times{p_2}timescdotstimes{p_n}$. Since each singleton is open, your topology is discrete.
edited Dec 18 '18 at 12:20
Henno Brandsma
112k348121
answered Dec 18 '18 at 12:10
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
add a comment |
$begingroup$
Your proof makes no sense, because the set that you defined makes no sense.
You can prove it as follows: is $p_1in X_1$, $p_2in X_2$, …, $p_nin X_n$, then each set ${p_i}$ is open and therefore, $bigl{(p_1,p_2,ldots,p_n)bigr}$ is an open subset of $prod_{i=1}^nX_i$, since it is equal to ${p_1}times{p_2}timescdotstimes{p_n}$. Since each singleton is open, your topology is discrete.
edited Dec 18 '18 at 12:20
Henno Brandsma
112k348121
answered Dec 18 '18 at 12:10
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
Your proof makes no sense, because the set that you defined makes no sense.
You can prove it as follows: is $p_1in X_1$, $p_2in X_2$, …, $p_nin X_n$, then each set ${p_i}$ is open and therefore, $bigl{(p_1,p_2,ldots,p_n)bigr}$ is an open subset of $prod_{i=1}^nX_i$, since it is equal to ${p_1}times{p_2}timescdotstimes{p_n}$. Since each singleton is open, your topology is discrete.
edited Dec 18 '18 at 12:20
Henno Brandsma
112k348121
answered Dec 18 '18 at 12:10
José Carlos SantosJosé Carlos Santos
168k22132236
edited Dec 18 '18 at 12:20
Henno Brandsma
112k348121
edited Dec 18 '18 at 12:20
Henno Brandsma
112k348121
edited Dec 18 '18 at 12:20
Henno Brandsma
112k348121
112k348121
answered Dec 18 '18 at 12:10
José Carlos SantosJosé Carlos Santos
168k22132236
answered Dec 18 '18 at 12:10
José Carlos SantosJosé Carlos Santos
168k22132236
answered Dec 18 '18 at 12:10
José Carlos SantosJosé Carlos Santos
168k22132236
168k22132236
add a comment |
add a comment |
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