Jtdylktuy

Let (X,d_1) and $(Y,d_2)$ be metric spaces. Further let $e$ be the metric in $Xtimes Y$ 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 $Xtimes Y$ 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:

$U_{x_1}=B(x_1,frac{epsilon}{2})={xin X:d_1(x_1,x)<frac{epsilon}{2}}$ for an arbitrary $x_1in X$

$V_{y_1}=B(y_1,frac{epsilon}{2})={yin Y:d_1(y_1,y)<frac{epsilon}{2}}$ for an arbitrary $y_1in Y$

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
$U_{x_1}times V_{y_1}=B(x_1,frac{epsilon}{2})times B(y_1,frac{epsilon}{2})=B((x_1,y_1,epsilon))={(x,yin Xtimes Y):e((x_i,y_i),(x,y))<epsilon}$

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?

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!