Let (X,d_1) and $(Y,d_2)$ be metric spaces. Further let $e$ be the metric in $X\times Y$ defined as $e((x_1,y_1),(x_2,y_2))=d_1(x_1,x_2)+d_2(y_1,y_2)$. Also let $\tau$ be the topology induced on $X\times Y$ by $e$. If $d_1$ and $d_2$ induces the topologies $\tau_1$ and $\tau_2$ on $X$ and $Y$, and respectively, and $\tau_2$ is the product topology of $(X,\tau_1)\times(Y,\tau_2)$ prove that $\tau=\tau_3$.

My proof:

Since $(X,d_1)$ and $(Y_2,d_2)$ are metric spaces the open sets are union of open balls such that:

$U_{x_1}=B(x_1,\frac{\epsilon}{2})=\{x\in X:d_1(x_1,x)<\frac{\epsilon}{2}\}$ for an arbitrary $x_1\in X$

$V_{y_1}=B(y_1,\frac{\epsilon}{2})=\{y\in Y:d_1(y_1,y)<\frac{\epsilon}{2}\}$ for an arbitrary $y_1\in Y$

The topology $\tau_1$ and $\tau_2$ are induced respectively by basis of open sets the form $U_{x_1}$ and $V_{y_1}$.

In the product topology space $(X,\tau_1)\times(Y,\tau_2)$ ,$U_{x_1}\times V_{y_1}$ are open sets that generate $\tau$, however $U_{x_1}\times V_{y_1}$ are in the topology $\tau_3$ once $U_{x_1}\times V_{y_1}=B(x_1,\frac{\epsilon}{2})\times B(y_1,\frac{\epsilon}{2})=B((x_1,y_1,\epsilon))=\{(x,y\in X\times Y):e((x_i,y_i),(x,y))<\epsilon\}$

once $e((x_i,y_i),(x,y))=d_1(x_1,x)+d_2(y_1,y)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$, which concludes the proof.


Is this proof right? If not. Why not? How should I correct it? Or provide an alternative one?

Thanks in advance!


That's a start, but there's more for you to do. Your argument (I believe) is that:

(1) $\{B(x_1, \epsilon/2) \times B(y_1, \epsilon/2) : x_1 \in X, y_1 \in Y, \epsilon > 0\}$ is a basis for $\tau_3$;

(2) $\{B((x_1, y_1), \epsilon) : x_1 \in X, y_1 \in Y, \epsilon > 0\}$ is a basis for $\tau$;

(3) these two bases are exactly the same, because $B(x_1, \epsilon/2) \times B(y_1, \epsilon/2) = B((x_1, y_1), \epsilon)$;

(4) therefore these two bases generate the same topology, which means $\tau = \tau_3$.

This would work except that (3) is false. For example, take $X$ and $Y$ both to be the real numbers with the usual metric, and draw $B(0, 1) \times B(0, 1)$ and $B((0, 0), 1)$: the first is a square and the second is a diamond around the square.

Your argument does show that $B(x_1, \epsilon/2) \times B(y_1, \epsilon/2) \subseteq B((x_1, y_1), \epsilon)$. That is, the second one can be larger, as in the example above. I think that will be useful.

Usually the best approach for showing that two sets are the same is to show that each contains the other, so I suggest you break the proof into two parts: show that (1) $\tau \subseteq \tau_3$, and (2) $\tau_3 \subseteq \tau$. To show (1), for example, it's enough to show that any basic open set $B((x_1, y_1), \epsilon)$ is in $\tau_3$; then all the other sets in $\tau$, which are unions of these basic open sets, must also be in $\tau_3$. This approach should also work for (2).

See if that helps!

  • $\begingroup$ Thanks for your answer. I cannot picture the diamond you talk. Could you provide some more insight into that particularity? Thanks in advance! $\endgroup$ Dec 25 '18 at 18:54
  • $\begingroup$ Sure. In that case $e$ is often called the "taxicab metric". There's more on that in Wikipedia, including a few examples. Try drawing a bunch of points, figure out which ones are inside $B((0, 0), 1)$ and which are not, and I think you will see the pattern. $\endgroup$
    – Hew Wolff
    Dec 26 '18 at 15:38

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