Here is exercise 6 from chapter 4:
If $f$ is defined on $E$, the graph of $f$ is the set of points $(x, f(x))$, for $x \in E$. In particular, if $E$ is a set of real numbers, and $f$ is real-valued, the graph of $f$ is a subset of the plane.
Suppose E is compact, and prove that f is continuous on E if and only if its graph is compact.
tl;dr – what should be the domain and codomain of $ f $ in this exercise?
Note that real numbers are mentioned as an example.
However other than that it seems to be implied that the proof should be carried out in a more general case?
If so, how general the proof should be?
I was trying to prove the statement assuming that the domain and codomain of $ f $ are arbitrary metric spaces.
However this seems too weak, because a graph of $ f $ is a subset of a Cartesian product of domain and codomain and it is not obviously clear how a distance function on a Cartesian product of two metric spaces should be defined.
Should I therefore limit myself to using $ \mathbb{R}^k $ and $ \mathbb{R}^s $ as domain and codomain? If so, should I then interpret the Cartesian product of those as simply $ \mathbb{R}^{k + s} $?
Best Answer
If $X_1$ and $X_2$ are metric spaces, then the metric defined on $X=X_1\times X_2$ by \begin{align*} d(x,y)=\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2} \end{align*} generates the product topology on $X$, under which $X$ is called a product space. It is not hard to verify that $d$ is indeed a metric. This is, of course, not the only metric that generates the product topology. Some other common choices include \begin{align*} d(x,y)=d_1(x_1,y_1)+d_2(x_2,y_2) \end{align*} and \begin{align*} d(x,y)=\max\{d_1(x_1,y_1),d_2(x_2,y_2)\}. \end{align*} It is a good exercise to verify that these definitions generate the same topology.
One may refer to any introductory topology textbook for more details on product spaces, in particular product spaces of metric spaces. To avoid digression, we simply point out that the usual topology on $\mathbb{R}^{k+s}$ and the product topology on $\mathbb{R}^k\times\mathbb{R}^s$ do agree.