Showing that two mappings are continuous. Introduction to topology

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.

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