Proof that $U\times V$ is open in the metric space $X\times Y$?

Question

Problem.

Let $(X,d_X)$ and $(Y,d_Y)$ be two metric spaces and let $U\subseteq X$ and $V\subseteq Y$ such that $U$ and $V$ are respectively open in $X$ and $Y$. Show that $U\times V$ is open in $(X\times Y,d_{X\times Y})$. Where $d_{X\times Y}((x_1,y_1),(x_2,y_2))=\max \{d_X(x_1,x_2),d_Y(y_1,y_2)\}$ for all $(x_1,y_1), (x_2,y_2)\in X\times Y$.

Proof.

Since $U\subseteq X$ and $U$ is open, for all $x\in U$ we can say that there exists $r_x>0$ such that the set $B_{d_X}(x,r_x)\subseteq U$ will be an open ball in $U$. Similarly the set $B_{d_Y}(y,r_y)\subseteq V$ will be an open ball in $V$.

Let $r=\min\{r_x,r_y\}$. Now we observe that,

\begin{align}(x_0,y_0)\in B_{d_{X\times Y}}((x,y),r)\cap (U\times V)&\implies d_{X\times Y}((x,y),(x_0,y_0))<r\\&\implies \max\{d_X(x,x_0),d_Y(y,y_0)\}<r\\&\implies (d_X(x,x_0)<r) \land (d_Y(y,y_0)<r)\\&\implies (x_0\in B_{d_X}(x,r))\land (y_0\in B_{d_Y}(y,r))\\&\implies (x_0,y_0)\in B_{d_X}(x,r)\times B_{d_Y}(y,r)\end{align}

So it follows that, $$B_{d_{X\times Y}}((x,y),r)\cap (U\times V)\subseteq B_{d_X}(x,r)\times (B_{d_Y}(y,r)\subseteq U\times V$$Now since $B_{d_{X\times Y}}((x,y),r)\cap (U\times V)$ is an open ball (this follows from the definition of an open ball) in $U\times V$ (it is easy to show that $B_{d_{X\times Y}}((x,y),r)\cap (U\times V)\subseteq U\times V$) we can say that $U\times V$ is open in $X\times Y$ and we are done.

Would there be any problem if we take $r=\max\{r_x,r_y\}$?

Answer 1

One problem with your proof is you did not tell us what definition of "open" you are using. There are several different definitions, the one usually used in metric spaces is:

$V\subset X$ is open if for every $x\in V$ there exists some $r$ such that $B(x,r)\subseteq V$.

Now if you are using this definition, you need to start with an arbitrary element $(x,y)\in U\times V$, and find the radius $r$ such that $B_{d_{X\times Y}}((x,y),r)$ is a subset of $U\times V$.

You didn't do that, so your proof is wrong, because you did not prove the inclusion at all. You did provide some $r$, but that $r$ is more or less arbitrary (since $r_x$ and $r_y$ are both arbitrary) and not good enough.

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Maybe you are using some other definition of open, but you need to tell us which one, because right now, I don't understand your proof. The statement $B_{d_{X\times Y}}((x,y),r)\cap (U\times V)\subseteq (U\times V)$ is trivially true, since $A\cap B\subset B$ is always true. Why would this imply that $U\times V$ is open?

What definition of "open" are you even using?

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After edit:

Your proof is still very confusing. See, you start with

"...Let $B_{d_X}(x,r_x)$ be an open ball in $X$...",

which implies that you already decided on what the value $r_x$ will be.

But then, you say

"...there exists $r_x>0$..."

which is confusing. You can't just now decide on a new value of $r_x$.

Furthermore, you say:

"exists $r_x$ such that $B(x,r_x)\cap U$ is an open ball in $U$"

This is true, however, by definition, this is true even if $U$ is not an open set!. What you really want is

"exists $r_x$ such that $B(x,r_x)$ is a subset of $U$."

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After second edit:

Still not good. You have proven that $B_{d_{X\times Y}}((x,y),r)\cap (U\times V)$ is a subset of $U\times V$, but that is trivially true (since, as I said, $A\cap B$ is always a subset of $B$), and that fact alone does not imply that $U\times V$ is open.

For example, $B(1,2)\cap [0,1]$ is a subset of $[0,1]$, but that does not mean that $[0,1]$ is an open set in $\mathbb R$.

What you need to prove is that there exists some ball in $X\times Y$ which:

1. Contains $(x,y)$
2. Is a subset of $U\times V$.

You still haven't done that.