## Open sets and closed sets in product topology

1. The problem statement, all variables and given/known data

Let $$(X_a, \tau_a), a \in A$$ be topological spaces, and let $$\displaystyle X = \prod_{a \in A} X_a$$.

2. Relevant equations

1. Prove that the projection maps $$p_a : X \to X_a$$ are open maps.

2. Let $$S_a \subseteq X_a$$ and let $$\displaystyle S = \prod_{a \in A} S_a \subseteq \prod_{a \in A} X_a$$. Prove that $$S$$ is closed iff each $$S_a \subseteq X_a$$ is closed.

3. Let $$T_a \subseteq X_a$$, prove that $$\displaystyle \overline{\prod_{a \in A} T_a} = \prod_{a \in A} \overline{T_a}$$.

4. If $$\abs{A} \leq \abs{\mathbb{N}}$$ and each $$X_a$$ is separable, prove that $$X$$ is separable.

3. The attempt at a solution

I don't know how to prove open/closed set problems in product topology. Can someone please give me some hint as to how I should approach these proofs? Some hints on each question will be even better.

1. This means that any open subset of the product space $$X$$ remains open when projected down to the $$X_\alpha$$.

Is it because the production topology $$\tau$$ for $$X$$ is the weakest topology with regard to $$\{ p_a :X \to X_a | a \in A \}$$?

2.

3.

4. $$X_a$$ is separable means there is a countable subset $$S_a \subseteq X_a$$ such that $$\overline{S_a} = X_a$$.
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 Write out what an open set in the product topology looks like: http://en.wikipedia.org/wiki/Product...ogy#Definition (the second paragraph). Based on this, where does the projection map send an arbitrary open set? For (4), use (3): take the product of the Sa and show that it's closure is X.

 Quote by VeeEight Write out what an open set in the product topology looks like: http://en.wikipedia.org/wiki/Product...ogy#Definition (the second paragraph). Based on this, where does the projection map send an arbitrary open set? For (4), use (3): take the product of the Sa and show that it's closure is X.

1. The product topology $$\tau$$ is generated from base $$\mathfrak{B}$$ consisting of product sets $$\displaystyle \prod_{a \in A} U_a$$ where only finitely many factors are not $$X_a$$ and the remaining factors are open sets in $$X_a$$. Therefore the project $$p_a$$ projects an open set $$S \subseteq X$$ to either $$X_a$$ or some open subset $$S_a \subset X_a$$.

2.

3.

4. $$X_a$$ is separable means there is a countable subset $$S_a \subseteq X_a$$ such that $$\overline{S_a} = X_a$$. Using previous result, we have
\begin{align*} \prod_{a \in A} \overline{S_a} = \prod_{a \in A} X_a = \overline{\prod_{a \in A} S_a} = X \end{align*}
Since $$S_a$$ is countable and $$|A| \leq |\mathbb{N}|$$, the cartesian product $$\displaystyle \prod_{a \in A} S_a$$ is countable. Hence $$X$$ is separable.

***This could be wrong, if $$|A| = |\mathbb{N}|$$, then the cartesian product does not have to be countable. So what is the set separable? Should the question say If $$|A| < |\mathbb{N}|$$ and each $$|X_a|$$ is separable, prove that $$|X|$$ is separable?