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I have to take an open set in $\mathbb{R}^2$ and show that it maps to an open set in $\mathbb{R}$.

So let $A \times B$ be an open set in $\mathbb{R}^2$. I have to show that $A$ is an open set.

By definiton of the product topology, an open set is a union of sets in the form $C \times D$ where $C$ and $D$ are both open in $\mathbb{R}$.

So $A \times B = (C \times D) \cup (E \times F) \cup (G \times H) ...$

But in general, $(C \times D) \cup (E \times F) \cup (G \times H) ... \neq (C \cup E \cup G ...) \times (D \cup F \cup H...) \neq A \times B$.

So I don't know how I can show that $A$ is an open set, if $(C \cup E \cup G ...) \neq A$.

mr eyeglasses
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2 Answers2

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Almost there. So not every (open) set in $\mathbb{R}^2$ can be decomposed as $A\times B$ for $A,B\subset \mathbb{R}$. But what you do know is that for $W\subset \mathbb{R}^2$ open, $W=\bigcup_{\alpha} U_{\alpha}\times V_{\alpha}$ for $U_{\alpha},V_{\alpha}\subset\mathbb{R}$ open and $\alpha$ some index set. So it suffices to show that $p(W)$ is open. But by general set theory, $$p(W)=p(\bigcup_{\alpha} U_{\alpha}\times V_{\alpha})=\bigcup_{\alpha} p(U_{\alpha}\times V_{\alpha})$$So it suffices to show that $p(U_{\alpha}\times V_{\alpha})$ is open for every $\alpha$. But this is obvious since $p(U_{\alpha}\times V_{\alpha})=U_{\alpha}$ is open.

Community
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Moya
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Let $U\subseteq\mathbb{R}^2$ be open. Suppose $p(U)$ is NOT open, so there is $x\in p(U)$ and a sequence $(x_n)\subseteq\mathbb{R}\backslash p(U)$ with $x_n\longrightarrow x$. Pick $(x, y)\in U$ with $p(x, y)=x$. Then all $(x_n, y)\notin U$, but $(x_n, y)\longrightarrow (x, y)\in U$. So $U$ is not open. Contradiction.

Damian Reding
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