Showing that two mappings are continuous. Introduction to topology

Question

I have a course were we are learning about topology and have to show in two different exercises that two different mappings are continuous. I think I have a solution. However, as these concepts a new to me, I would like to ask if this is indeed correct and if there is a better way to solve them.

The first problem: Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and define $d_{X\times Y}:(X\times Y)\times (X\times Y) \rightarrow \mathbb{R}^+_0$ by

$d_{X\times Y}((x_1,y_1),(x_2,y_2))=max(d_X(x_1,x_2),d_Y(y_1,y_2))$

Then show the projection

$p_X:X\times Y \rightarrow X, p_X(x,y)=x$

is continuous.

I would then solve this by showing that for every open set V in X then $p_X^{-1}(V)$ is an open set in $X\times Y$. So in mathematical terms

$V \subseteq X$ and

$\forall x\in V, \exists \delta_V >0 : B^X_{\delta_V}(x)\subseteq V$ and $B^X_{\delta_V}(x)=\{y\in X|d_X(x,y)<\delta_V\}$

We then need there to exist a $\delta_{X\times Y}$ so that $\forall (x,y)\in p_X^{-1}(V), \exists \delta_{X\times Y} >0 : B^{X\times Y}_{\delta_{X\times Y}}((x,y))\subseteq p_X^{-1}(V)$

Where $p_X^{-1}(V)=V\times Y$ as the second coordinate in $(x,y)$ is just dropped and can therefore be anything.

As every $x_1 \in V$ has a value $\delta_X$ we can now look at every $(x_1,y)\in p_X^{-1}(V)$, where it is the same element $x_1$ as before, then $B^{X\times Y}_{\delta_{X\times Y}}((x_1,y))\subseteq p_X^{-1}(V)$ if $0<\delta_{X\times Y} \leq \delta_X$.

I conclude this as for $B^{X\times Y}_{\delta_{X\times Y}}((x,y))\nsubseteq p_X^{-1}(V)$ then there has to be a point $(x_2,y_2)$ where $d_X(x_1,x_2)<\delta_{X\times Y}\leq\delta_X$ and $x_2\notin V$ as every possible $y_2$ is in $p_X^{-1}(V)=V\times Y$. But this is contradictory to what we know as $\delta_X$ is chosen so every $x_2$ where $d_X(x1,x2)<\delta_X$ is in V.

We can therefore conclude that for every open subset of X, called $V$, then $p_X^{-1}(V)$ is an open subset of $X \times Y$ as $\forall (x,y)\in p_X^{-1}(V), \exists \delta_{X\times Y} >0 : B^{X\times Y}_{\delta_{X\times Y}}((x,y))\subseteq p_X^{-1}(V)$ is fulfilled and the projection is therefore continuous.

The second problem is so similar that it is not currently added.

Answer 1

It suffices to show that for any $(x,y)\in p_X^{-1}V$ there exists an open ball $A=B_{d_{X\times Y}}((x,y),r)$ centered at $(x,y),$ with $r>0,$ such that $A\subseteq p_X^{-1}V.$

For any $x'\in X$ we have $p_X^{-1}\{x'\}=\{x'\}\times Y.$ So for any $U\subset X $ we have $p_X^{-1}U=\cup_{x'\in U}(\{x'\}\times Y)=U\times Y.$

So if $(x,y)\in p_X^{-1}V=V\times Y$ then $x\in V,$ and since $V$ is open in $X,$ there exists $r>0$ such that $B_{d_X}(x,r)\subset V.$ Therefore $$ B_{d_{X\times Y}}((x,y),r)=B_{d_X}(x,r)\times B_{d_Y}(y,r)\subseteq$$ $$\subseteq B_{d_X}(x,r)\times Y\subseteq$$ $$\subseteq V\times Y=$$ $$=p_X^{-1}V.$$

Answer 2

The projection p:X×Y -> X, (x,y) -> x is continuous.
Proof. Let U be an open subset of X.
Then p$^{-1}$(U) = U×Y is open subset of X×Y
since it is the product of two open sets.

That is true of all topological spaces.
For your problem simply show that the function
d = d$_{X×Y}$ is a metric and U×Y is open when U is.

Actually, the fruitful proposition is to prove is that d
generates the product topology - whenever U is open within
X and V is open within Y, then U×V is open within X×Y.