Mathematics III - Foundations of Analysis - Summer 24

Many results on continuous functions from a metric space to another metric space are similar to those from a subset of the reals to another subset of the reals.  We simply use the metric in places of the distance normally defined by the absolute value.

Here is an example.  Let $f$ from the metric space $X$ to the metric space $Y$ be continuous at a point $p\in X$. Then for any $\varepsilon>0$ there is  $\delta>0$ as in the definition of continuity.  Since $p_n\to p$ we can choose some $N$ such that $d_X(p_n,p)$ for all $n\ge N$, and this means that  $d_Y(f(p_n),f(p))$ for all $n\ge N$.  Hence $f(p_n)\to f(p).$

We can also prove the converse statement.  To this end we assume that $f(p_n)\to f(p)$ but $f$ is not continuous at $p$.  The latter means that we can find some $\varepsilon>0$ and some $p_n\in X$ such tha $d_Y(f(p),f(p_n))\ge\varepsilon$ yet $d_X(p_n,p)$.  Clearly this is a contradiction. 

So we have proved a useful theorem concerning continuity. 

Theorem. A function $f$ from a metric space $X$ to a metric space $Y$ is continuous at a point $p\in X$ if and only if $\displaystyle\lim_{n\to\infty}f(p_n)=f(p)$ for every sequence $\{p_n\}$ which converges to $p$.

In fact there is an equivalent definition of a continuous function of a metric space (often used in topology) due to the following theorem

Theorem. A mapping $f$ of a metric space $X$ into a metric space $Y$ is continuous on $X$ if and only if $f^{-1}(V)$ (the inverse image) is open in $X$ for every open set $V$ in $Y.$

There is a discussion on how to write the proof and here.   What do you think?

Since a set is closed if and only if its complement is open and $f^{-1}(E^c)=(f(^{-1}(E))^c$ for every $E\subset Y$ we have

Corollary. A mapping $f$ of a metric space $X$ into a metric space $Y$ is continuous on $X$ if and only if $f^{-1}(C)$ (the inverse image) is closed in $X$ for every closed set $C$ in $Y.$

Caution: The continuity of $f$ is not equivalent to the requirement that the image $f(U)$ of an open set $U\subset X$ be open in $Y$. An example is $f:{\mathbb R}\to{\mathbb R}$, $f(x)=|x|.$  Here $f({\mathbb R})=\{x\in{\mathbb R}:x\ge0\}$, which is not open in ${\mathbb R}$.

Example.

f(x) = x²

First, let's look at the inverse images of an open interval:

• If $0$ then the inverse image of the open interval $(a,b)$ is $(-\sqrt{b},-\sqrt{a})\cup(\sqrt{a},\sqrt{b})$.  In particular, the inverse image is open.
• If $a$, then the inverse image of the open interval $(a,b)$ is $(-\sqrt{b},\sqrt{b})$, which is again open.
• If $a$, then the inverse image of the open interval $(a,b)$ is again open (which set is it ?)

But it is now obvious that the inverse image of closed intervals is again a closed set (note that the empty set is both open and closed).

Hence, we have proved that the function $f(x)=x^2$ is continuous, avoiding the tedious epsilon-delta proof.  ♦

Now we use the theorem to prove the composition function of two continuous functions is continuous.  More precisely,

Theorem.  Suppose that $X,Y,Z$ are metric spaces, $E\subset X$, $f$ maps $E$ into $Y$, $g$ maps the range of $f$, $f(E)$ into $Z$, and $h$ is the mapping of $E$ into $Z$ defined by $h(x)=g(f(x))$ ($x\in E$).  If $f,g$ are continuous on $E$ and $f(E)$ respectively, then $h$ is continuous. 

Proof.  Let $V\subset Z$ be open. The key observation here is $h^{-1}(V)=f^{-1}(g^{-1}(V))$. (This can be proven by definition.)  But $g^{-1}(V)=f(E)\cap U$ is open for some open set $U$ in $Y$.  Then $f^{-1}(g^{-1}(V))=f^{-1}(f(E)\cap U)=f^{-1}(U)$ is open in $E$, since $f$ and $g$ are continuous.  Therefore $h$ is continuous. ◊