Exercise 2.16

Proof. Choose a sequence $x_{n}n$ whose limit is an arbitrary $a$. By compactness of $K$, the graph $G$ of $f$ contains a subsequence $(x_{\rho (n)},f(x_{\rho (n)}))$ of $(x_n,f(x_n))$ that converges to some $(a,b)$ of $X\times K$. $G$ is closed; therefore, $(x_{\rho (n)},f(x_{\rho (n)}))$ converges in $G$. So, $b=f(a)$; which establishes that $f$ is sequentially continuous. Since $X$ is metrizable, $f$ is also continuous; see [A6]. So ends the proof.

To show that compactness cannot be omitted from the hypotheses, we showcase the following counterexample,

Clearly, $f$ has a discontinuity at $0$. Nevertheless the graph $G$ of $f$ is closed. To see that, first remark that

Next, let $(x_n,1∕x_n)$ be a sequence in $G_{\text{+}}=(x,1∕x)x>0$ that converges to $(a,b)$. To be more specific: $a=0$ contradicts the boundedness of $(x_n,1∕x_n)$: $a$ is necessarily positive and $b=1∕a$, since $x\mapsto 1∕x$ is continuous on $R_{\text{+}}$. This establishes that $(a,b)\in G_{\text{+}}$, hence the closedness $G_{\text{+}}$. Finally, we conclude that $G$ is closed, as a finite union of closed sets. □