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Let $X=X_1$ x $X_2$ where $(X_1,d_1)$ and $(X_2,d_2)$ are metric spaces. Equip $X$ with a product metric $d$. Define a map $\Pi_1:X \to X_1$ by $\Pi_1(x_1,x_2) = x_1$. Let $U \subset X$ be open and let $x_0 \in U_1 = \Pi_1(U)$. As $x_0 \in U_1 \exists y_0 \in X_2$ such that $(x_0,y_0) \in U$. As $U$ is open $\exists \delta>0$ such that $B_{\delta}((x_0,y_0)) \subset U$. As $\Pi_1$ is continuous, $\Pi_1(B_{\delta}((x_0,y_0))) = B_{\epsilon}(x_0)$ and we have that $B_{\epsilon}(x_0) \subset U_1$.

I was unsure if this was correct as this would imply that $B_{\epsilon}(x_0) \subset U_1 \forall \epsilon>0$, which can't be true but couldn't see where the proof breaks down.

user60327
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1 Answers1

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Let $U \subset X_1 \times X_2$ be open, we have that $U$ can be expressed as $\bigcup_i A_i \times B_i$ where $A_i \subseteq X_1$ is open and $B_i \subseteq X_2$ is open, then

$$\pi_1 (U) = \pi_1 \big(\bigcup_i A_i \times B_i \big) = \bigcup_i \pi_1 \big(A_i \times B_i \big) = \bigcup_i A_i$$

aram
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  • I know that this is a proof but wondered where my proof broke down. – user60327 Nov 16 '14 at 22:27
  • @Emma Well I think you're missing one step, and that would be saying that the open ball is the product of open balls in each space, then the projection gives you the open ball in $X_1$. – aram Nov 16 '14 at 22:38