Lemma 8

Let X and Y be metric spaces, the space X being complete. Let $f$ be a mapping from X into Y such that

for every $x$ and $y$ in X and such that the graph of $f$ is closed in $X\times Y$. Then $f$ is a closed mapping.

Let F be a closed subset of X and let $(y_n)_{n\in\mathbf{N}}$ be a sequence in $f(F)$ which converges to $y\in Y$. For every $n\in \mathbf{N}$, let $x_n\in F$ be such that $f(x_n) =y_n$. The assumption implies that the sequence $(x_n)_{n\in\mathbf{N}}$ is a Cauchy sequence; let $x\in X$ be its limit; it is an element of F, since F is closed. Moreover, one has $(x_n, f(x_n))\rightarrow (x, y)$ in $X\times Y$. Since the graph of $f$ is closed, it follows that $y=f(x)$ belongs to $f(F)$.