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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 $d_{Xtimes Y}:(Xtimes Y)times (Xtimes Y) rightarrow mathbb{R}^+_0$ by

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

Then show the projection

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

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 $Xtimes Y$.
So in mathematical terms

$V subseteq X$ and

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

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

Where $p_X^{-1}(V)=Vtimes Y$ 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 $B^{Xtimes Y}_{delta_{Xtimes Y}}((x_1,y))subseteq p_X^{-1}(V)$ if $0<delta_{Xtimes Y} leq delta_X$.

I conclude this as for $B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))nsubseteq p_X^{-1}(V)$ then there has to be a point $(x_2,y_2)$ where $d_X(x_1,x_2)<delta_{Xtimes Y}leqdelta_X$ and $x_2notin V$ as every possible $y_2$ is in $p_X^{-1}(V)=Vtimes Y$. 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 $forall (x,y)in p_X^{-1}(V), exists delta_{Xtimes Y} >0 : B^{Xtimes Y}_{delta_{Xtimes Y}}((x,y))subseteq p_X^{-1}(V)$ is fulfilled and the projection is therefore continuous.

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

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

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

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

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.