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$(X, \mathcal{T}_X), (Y,\mathcal{T}_Y)$ be topological space, and let $A, B$ be subset of $X,Y$ respectively.

Now, $(X \times Y, \mathcal{T}_{X \times Y})$ be product topology. $A \times B \subset X \times Y$. Consider subspace topology $\mathcal{T}_1$ on $A \times B$

Also, $(A, \mathcal{T}_A), (B, \mathcal{T}_B)$ where $\mathcal{T}_A$ and $\mathcal{T}_B$ are subspace topologies on $A, B$ respectively,consider product topology $\mathcal{T}_2$ on $A \times B$

Prove $\mathcal{T}_1 = \mathcal{T}_2$

My idea is to show that the bases of $\mathcal{T}_1$ and $\mathcal{T}_2$ are same. where $\mathcal{T}_1$ basis is {$(A \times B) \cap (U \times V) | (U \times V) \in \mathcal{T}_{X \times Y}$}

and $\mathcal{T}_2$ basis is {$U_0 \times V_0 | U_0 \in \mathcal{T}_A, V_0 \in \mathcal{T}_B$}. I know how to show these two bases are equal. However, I don't if this is sufficient to prove two topologies are equal.

ElleryL
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  • In general yes. But for this question it isn't really necessary. – angryavian Jun 25 '15 at 02:26
  • That should work. A basis generates a topology. If that is not convincing enough you could suppose one topology has an element that the other does not have. You should be able to dismiss that case with a simple contradiction. – graydad Jun 25 '15 at 02:26
  • @angryavian could u tell me other ways of solving this question? – ElleryL Jun 25 '15 at 02:31
  • @graydad, so it's true that a basis generating an unique topology? – ElleryL Jun 25 '15 at 02:31
  • Should be true! I haven't rigorously proved it, but if you think about how the basis generates a topology, it doesn't seem possible that there could be more than one. You could probably prove this result too if you are unsure. – graydad Jun 25 '15 at 02:33

1 Answers1

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It is sufficient, and here is why.

Let $(X,\mathcal{T}_1)$ and $(Y,\mathcal{T}_2)$ be topological spaces with the same basis $\mathcal{B}$. For the sake of contradiction, suppose $\mathcal{T}_1 \neq \mathcal{T}_2$. Then $\mathcal{T}_1$ contains an element $V$ such that $V\notin \mathcal{T}_2$. Since $V \in \mathcal{T}_1$ there exists a collection of basic elements $\{B_\alpha \}_{\alpha \in \lambda} \subseteq \mathcal{B}$ such that $$V = \bigcup_{\alpha \in \lambda} B_\alpha $$ And since $\mathcal{B}$ is a basis for $\mathcal{T}_2$ then $\bigcup_{\alpha \in \lambda} B_\alpha$ is an element of $\mathcal{T}_2$, a contradiction.

graydad
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    Does the converse hold as well, i.e. if they are the same topologies, then they have the same base? Cheers! – asn32 Mar 18 '18 at 18:33