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If $(X, d_x)$ and $(Y, d_Y)$ are connected, show that $(X \times Y, d)$ where $d((x_1, y_1), (x_2, y_2)) = d_x(x_1, x_2) + d_Y(y_1, y_2)$ is connected.

Proof: Suppose that $(X \times Y, d)$ is not connected, then there exist open sets $U$, $V$ of $(X \times Y, d)$ such that $U \cup V = X \times Y$ and $U \cap V = \emptyset$ and $U \neq \emptyset$ and $V \neq \emptyset$

We claim that $\pi_X : (X \times Y, d) \to (X, d_X)$ defined by $\pi_X(x, y) =x$ is an open map.

Firstly recall that $\pi_x[U] = \{\pi_X(x, y) \ | \ (x, y) \in U\}$.

Pick $x \in \pi_X[U]$, then $x = \pi_X(x, y)$ for some $(x, y) \in U$. Since $U$ is open in $(X \times Y, d)$ there exists an $r > 0$ such that $B_{(X \times Y, d)}\left((x, y), r\right) \subseteq U$. We will now show that $B_{(X, d_X)}(x, r) \subseteq \pi_X[U]$.

Let $\alpha \in B_{(X, d_X)}(x, r)$, then $d_X(x, \alpha) < r$. Now note that $(\alpha, y) \in B_{(X \times Y, d)}\left((x, y), r\right)$ since $$d\left((x, y), (\alpha, y)\right) = d_X(x,\alpha) + d_Y(y, y) = d_X(x, \alpha) + 0 = d_X(x, \alpha) < r$$

Then we have $\pi_X(\alpha, y) = \alpha$ hence $\alpha \in \pi_X[U]$, thus proving that $B_{(X, d_X)}(x, r) \subseteq \pi_X[U]$. Since $x$ was chosen arbitrarily we have $\pi_X[U]$ to be open in $(X, d_X)$. Therefore $\pi_X$ is an open map.

Hence since $U$ and $V$ are open sets of $(X \times Y, d)$ we have $\pi_X[U]$ and $\pi_X[V]$ to be open sets of $X$.

Now it is relatively straightforward to show that $\pi_X[U] \cup \pi_x[V] = X$ and since $U \cap V = \emptyset$ we have $\pi_X[U] \cap \pi_X[V] = \emptyset$. Furthermore we have $\pi_X[U] \neq \emptyset$ since $U \neq \emptyset$ and $\pi_X[V] \neq \emptyset$ since $V \neq \emptyset$. Thus $X$ is disconnected a contradiction. $\square$


Is my proof correct? Can I improve it in any way? Are there any shorter or more efficient ways to prove this?

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  • $\begingroup$ Your proof is not correct. $U\cap V=\emptyset$ does not imply that $\pi_X(U)\cap\pi_X(V)=\emptyset$. The simpliest example is $X=Y=\{1,2\}$, $U=\{(1,1)\}$ and $V=X^2\backslash\{(1,1)\}$ with discrete metric on both $X,Y$. $\endgroup$ – freakish May 1 '18 at 12:06
  • $\begingroup$ @freakish Is the part of my proof that shows that $\pi_X$ is an open map correct? $\endgroup$ – Perturbative May 1 '18 at 12:45
  • $\begingroup$ It seems to be correct. $\endgroup$ – freakish May 1 '18 at 18:10
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i) Consider a subset $A_{xy}:=x\times Y\cup X\times y$. Then $$A_{xy}\cap A_{ab}\neq \emptyset $$

Then $X\times Y=A_{xy}\cup (\bigcup_{a,b} A_{ab})$ is connected if $A_{xy}$ is connected.

ii) $i_X :(X,d_x)\rightarrow (X\times Y,d)$ is continuous so that $i_X(X)$ is connected.

So $i_Y(Y)$ is connected. Since $i_X(X)\cap i_Y(Y)\neq \emptyset $, then $A_{xy}$ is connected.

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Your proof is not correct. You claim that $\pi_X(U)\cap\pi_X(V)=\emptyset$. This is not true in general. Suppose that $X=Y=\mathbb R$, that $U=\{(x,x)\,|\,x\in\mathbb{R}\}$ and that $V=\mathbb{R}^2\setminus U$. Then $U\cap V=\emptyset$, but $\pi_X(U)=\pi_X(V)=\mathbb R$. However, if you consider the discrete metric in $\mathbb R$, both $U$ and $V$ will be open.

You will find a proof here.

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  • $\begingroup$ Except that $U$ is not open in your case. $\endgroup$ – freakish May 1 '18 at 12:01
  • $\begingroup$ @freakish I know that. But at no point the fact that $U$ is open was used in the proof. $\endgroup$ – José Carlos Santos May 1 '18 at 12:02
  • $\begingroup$ It was assumed by the OP. To find a proper counterexample you need to show that it fails when $U$ is open. See my comment to the question. $\endgroup$ – freakish May 1 '18 at 12:03
  • $\begingroup$ @freakish Note that $U$ is open if I consider in $\mathbb R$ the discrete topology. $\endgroup$ – José Carlos Santos May 1 '18 at 12:04
  • $\begingroup$ Fair enough. You may want to update your answer. $\endgroup$ – freakish May 1 '18 at 12:07

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