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I wish to prove the following statement. Let $X$ and $Y$ be topological spaces with $A \subset X$ and $B\subset Y$. Then prove that the topology on $A \times B$ as a subspace of the product $X\times Y$ is the same as the product topology on $A\times B$, where $A$ and $B$ have the subspace topology from $X$ and $Y$ respectively.

Well I understand that the first thing looks like:

$$ \{(A\times B) \cap (U_i \times V_i)|U_i,V_i \in \text{topology on $X$ and $Y$ respectively}\} $$

Isn't it a set theory law that this is $$ \{(A\cap U_i) \times (B \cap V_i)|U_i,V_i \in \text{topology on $X$ and $Y$ respectively}\} $$ Then we are done. Or do I have to use double containment? Also, does anyone have a good list of the non-special case set distributive laws?

Thank you.

janmarqz
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Riemann-bitcoin.
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    Your first expression seems to assume that the set ${U_i \times V_i ,\bigm|, U_i,V_i \in \text{topology on $X$ and $Y$ respectively}}$ is the product topology on $X \times Y$, but that is false. That set is only a basis for the product topology. – Lee Mosher Oct 31 '16 at 12:46
  • Yes but I was meaning the { } to indicate a basis.does this not work? – Riemann-bitcoin. Oct 31 '16 at 12:47
  • Since you did not say that the ${}$ indicates a basis, and instead said that ${}$ indicates the topology, then no, it does not work. If you do decide to say that the ${}$ indicates a basis, then you will have to correctly apply theorems which relate bases and the topologies that they generate, because what you must prove is equality of the topologies. – Lee Mosher Oct 31 '16 at 12:52

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