# Proving the Product of topological spaces are metrizable

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.

Question:

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

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!

• Thanks for your answer. I cannot picture the diamond you talk. Could you provide some more insight into that particularity? Thanks in advance! Dec 25 '18 at 18:54
• 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. Dec 26 '18 at 15:38