Bonjour, I'm doing Problem III.2.18 from textbook Analysis I by Amann/Escher.
$p$ is open if the images of open sets under $p$ are open.
$X \times Y$ is endowed with the product metric.
Could you please verify if my proof look fine or contains logical gaps/errors? Any suggestion is greatly appreciated.
My attempt:
If $(X_{k}, d_{k})$ are metric spaces for $1 \leq k \leq m$, then $X = \prod_{k=1}^{m} X_k$ is a metric space with respect to the product metric $d:X \times X \to \mathbb R$ where $$d(x, y) =\max_{1 \leq k \leq m} d_{k}(x_{k}, y_{k})$$ for $x =(x_{1}, \ldots, x_{m}) \in X$ and $y =(y_{1}, \ldots, y_{m}) \in X$.
Lemma 1: The canonical projections $$\operatorname{pr}_{k} : X \rightarrow X_{k}, \quad x=(x_{1}, \ldots, x_{m}) \mapsto x_{k}, \quad 1 \leq k \leq m$$ are Lipschitz continuous.
Lemma 2: $$\mathbb{B}_{X}(x, r) = \prod_{k=1}^{m} \mathbb{B}_{X_{k}}(x_k, r), \quad \overline{\mathbb{B}}_{X}(x, r) = \prod_{k=1}^{m} \overline{\mathbb{B}}_{X_{k}} (x_k, r)$$ for all $x=(x_{1}, \ldots, x_{m}) \in X$ and $r>0$.
- $p$ is continuous
It follows directly from Lemma 1 that $p$ is Lipschitz continuous and thus continuous.
- $p$ is open
Approach 1:
Assume that $A$ is an open subset of $X \times Y$ and $A_1 = p[A] \subseteq X$. For $x_0 \in A_1$, there is $y_0 \in Y$ such that $(x_0, y_0) \in A$. Because $A$ is open in $X \times Y$, there is $r>0$ such that $\mathbb B_{X \times Y} ((x_0, y_0),r) \subseteq A$. Then $p[\mathbb B_{X \times Y} ((x_0, y_0),r)] \subseteq p[A] = A_1$. We have $$\begin{aligned} x \in \mathbb B_X (x_0,r) &\implies d_X(x,x_0) < r \\ &\implies \max \{d_X(x,x_0), d_Y(y_0,y_0)\} < r \\ &\implies d_{X \times Y} ((x, y_0),(x_0, y_0)) < r \\& \implies (x, y_0) \in \mathbb B_{X \times Y} ((x_0, y_0),r)\\ &\implies x \in p[\mathbb B_{X \times Y} ((x_0, y_0),r)]\\ &\implies x \in A_1\end{aligned}$$
As such, $\mathbb B_X (x_0,r) \subseteq A_1$. Hence $A_1$ is open in $X$.
Approach 2:
Assume that $A$ is an open subset of $X \times Y$ and $A_1 = p[A] \subseteq X$. For $x_0 \in A_1$, there is $y_0 \in Y$ such that $(x_0, y_0) \in A$. Because $A$ is open in $X \times Y$, there is $r>0$ such that $\mathbb B_{X \times Y} ((x_0, y_0),r) \subseteq A$. Then $p[\mathbb B_{X \times Y} ((x_0, y_0),r)] \subseteq p[A] = A_1$. It follows from Lemma 2 that $$\begin{aligned} p[\mathbb B_{X \times Y} ((x_0, y_0),r)] = p \left[ \mathbb{B}_{X}(x_0, r) \times \mathbb{B}_{Y}(y_0, r) \right] = \mathbb{B}_{X}(x_0, r) \end{aligned}$$
As such, $\mathbb B_X (x_0,r) \subseteq A_1$. Hence $A_1$ is open in $X$.
Yes, you proofs are correct.
You did not prove that $p$ is, in general, not closed. Just take $X=Y=\mathbb R$, with the usual metric. Then the set $\left\{(x,y)\in\mathbb R^2\,\middle|\,y=\frac1x\right\}$ is closed, but its projection is $\mathbb R\setminus\{0\}$, which is not.