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Let $E_1$ and $E_2$ metric spaces and $E=E_1\times E_2$ a metric spaces with some metric $d$. Let $\pi_1$ and $\pi_2$ the projections maps of $E_1\times E_2\rightarrow E_1$ and $E_1\times E_2\rightarrow E_2$ respectly, i.e, $$\pi_1(x, y)=x,\,\,\,\,\,\,\,\,\,\pi_2(x, y)=y;\,\,\,\,\,\,\,\,\forall\,\,(x, y)\in E_1\times E_2$$

I know that if $A\subseteq E$ is open then $\pi_1(A)$ is also open in $E_1$ and $\pi_2(A)$ is also open in $E_2$, but the reverse is true? i.e if $\pi_1(B)$ is open in $E_1$ and $\pi_2(B)$ is open in $E_2$ then $B$ is open in $E_1\times E_2$?

Roiner Segura Cubero
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2 Answers2

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No, that's not true. A slight variation on Cameron Buie's example: let $E_1=E_2=\mathbb{R}$ and let $$A=\{(x,0):x\in (-1,1)\}\cup\{(0,y):y\in(-1,1)\},$$ i.e., a "plus sign" in the plane $\mathbb{R}^2$. Then, projected in either direction onto $\mathbb{R}$, we get $(-1,1)$ which is open, but $A$ itself is not open in $\mathbb{R}^2$.

Zev Chonoles
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No. Consider $E_1=E_2=\Bbb R,$ and let $B$ be the open unit disk about the origin in $\Bbb R^2$ with the point $\left\langle\frac{\sqrt2}2,\frac{\sqrt2}2\right\rangle$ added. Then $B$ is not open in $\Bbb R^2$, but $\pi_1(B)=\pi_2(B)=(-1,1)$ is open in $\Bbb R$.

Added: We can say (for example) that if $C\subseteq E_1$ is open, then the preimage $\pi_j^{-1}(C)$ is open in $E$, since $\pi_1^{-1}(C)=C\times E_2$ for any $C\subseteq E_1.$ However, while we have in general that $$\pi_1^{-1}\bigl(\pi_1(B)\bigr)\supseteq B$$ for any $B\subseteq E,$ we usually don't have equality, so the converse you mentioned doesn't follow from this fact.

Cameron Buie
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  • Cameron as you prove that if $A$ is open in $E$ then $\pi_1(A)$ is open in $E_1$? – Roiner Segura Cubero Jun 22 '13 at 17:29
  • No problem. Hopefully it's clear to you what's going on in the three examples given here. The projection maps basically "cast a shadow" of the sets onto their respective axes. This shadow will "look like" an open set to the axis whenever the set itself is open--that is, for example, $\pi_1(A)$ will be open in $E_1$ whenever $A$ is open in $E$--but these examples suggest that there are many different sets that "cast the same shadows," and not all of them are open, even if the shadows are. – Cameron Buie Jun 22 '13 at 17:32
  • Adapting David's example, take the open line segment in the plane from $\langle-1,-1\rangle$ to $\langle 1,1\rangle$. This will give the same shadow as the examples Zev and I gave. – Cameron Buie Jun 22 '13 at 17:35
  • Thanks Cameron I were clear examples. But true that if $A$ is open in $E$ then $\pi_1(A)$ is open in $ E_1 $? My question is how would you do this test? – Roiner Segura Cubero Jun 22 '13 at 17:38
  • As for proving the other direction, what do you know about how the product topology is defined? Depending on the definition, the proof may be almost trivial. – Cameron Buie Jun 22 '13 at 17:39
  • DEFINITION OF OPEN SETS IN $E_1\times E_2$ A set $A \subseteq E_1\times E_2$ is called open in $E_1\times E_2$ iff it is the union of Cartesian products $G\times H$, where $G$ is open in $E_1$ and $H$ is open $E_2$ – Roiner Segura Cubero Jun 22 '13 at 17:51
  • Exactly. What would $\pi_1(G\times H)$ be? If $\mathcal A$ is any collection of subsets of $E$ and $A=\bigcup_{B\in\mathcal A}B,$ what can you say about the relationship between $$\bigcup_{B\in\mathcal A}\pi_1(B)$$ and $\pi_1(A)$? – Cameron Buie Jun 22 '13 at 18:03