#ref(remark)
#tex(impre) \begin{eqnarray*} y\in f(A) & \qquad\Longleftrightarrow \qquad & \mbox{there is an $x\in A$ such that $y=f(x)$}\\ x\in f^{-1}(B) & \qquad\Longleftrightarrow\qquad & f(x)\in B \end{eqnarray*} #end #ref(impre)
#ref(line)
#ref(example)
Consider the function
#tex(ex11) $f(x)=x^2$ #end #ref(ex11)
which is a map of the real line to itself. Then:
#tex(ex12) {\bf (i)} $f^{-1}(\;(1,4)\;) = (-2,-1)\cup(1,2)$ #end #ref(ex12)
#link(Ex12)#ref(question)#end Prove it.
#ref(line)
#sideline(Ex12) #ref(ex12)
#ref(line)
#tex
\begin{eqnarray*}
x\in f^{-1}(\;(1,4)\;) & \qquad\Longleftrightarrow \qquad & f(x)=x^2\in(1,4) \\
& \qquad\Longleftrightarrow \qquad & 1< x^2 < 4 \\
& \qquad\Longleftrightarrow \qquad & -2
#tex(ex13)
{\bf (ii)}
$f^{-1}(\;(-5,-3]\;) = \emptyset$
#end
#ref(ex13)
#link(Ex13)#ref(question)#end
Prove it.
#ref(line)
#sideline(Ex13)
#ref(ex13)
#ref(line)
#tex
\begin{eqnarray*}
x\in f^{-1}(\;(-5,-3]\;) & \qquad\Longleftrightarrow \qquad & f(x)=x^2\in(-5,-3] \\
& \qquad\Longleftrightarrow \qquad & -5< x^2 \le -3 \\
\end{eqnarray*}
This latter condition can clearly never happen for any $x\in \R^2$.
#end
#end
#tex(ex14)
{\bf (iii)}
$f^{-1}(\;\{25\}\;) = \{-5,5\}$
#end
#ref(ex14)
#link(Ex14)#ref(question)#end
Prove it.
#ref(line)
#sideline(Ex14)
#ref(ex14)
#ref(line)
#tex
\begin{eqnarray*}
x\in f^{-1}(\;\{25\}\;) & \qquad\Longleftrightarrow \qquad & f(x)=x^2\in\{25\} \\
& \qquad\Longleftrightarrow \qquad & x^2=25 \\
& \qquad\Longleftrightarrow \qquad & x=\pm 5 \\
& \qquad\Longleftrightarrow \qquad & x\in\{-5,5\}
\end{eqnarray*}
#end
#end
#tex(ex15)
{\bf (iv)}
$f^{-1}(\;25\;)$ doesn't mean anything.
#end
#ref(ex15)
#link(Ex15)#ref(question)#end
Explain this.
#sideline(Ex15)
#ref(ex15)
The quantity inside the parentheses is not a set. This preimage is only
defined on sets of real numbers.
#end
#ref(line)
Now we come to the characterization of continuity in terms of open sets.
#tex(prp514)
{\bf Proposition 5.14.}
Let $f:X\rightarrow Y$ be a map between the metric spaces $(X,\rho)$ and
$(Y,\sigma)$. Then $f$ is continuous if and only if $f^{-1}(U)$ is open
for all open set $U\subseteq Y$.
#end
#ref(prp514)
#link(Prp514)Proof.#end
#sideline(Prp514)
#ref(prp514)
#ref(line)
#tex
{\bf Proof.}
Suppose first that $f$ is continuous, and show that whenever $U\subseteq
Y$ is open, its preimage, $f^{-1}(U)$ is also open. To show that $f^{-1}(U)$
is open, we must pick an arbitrary point $x_0\in f^{-1}(U)$ and find an
$r>0$ such that $N_\rho(x_0, r)\subseteq f^{-1}(U)$ (we put subscripts
on $N_\rho(x,r)$ and $N_\sigma(y,s)$ to emphasize that the open
balls are defined in terms of different metrics).
\vskip5pt
\noindent
Since $x_0\in f^{-1}(U)$, this
means that $f(x_0)\in U$. Since $U$ is open, there is an $\eps>0$ such
that $N_\sigma(f(x_0), \eps)\subseteq U$. Since $f$ is continuous, it is
continuous at $x_0$, and so there is a $\delta>0$ such that
$$\sigma(f(x),f(x_0))<\eps\mbox{ for all $x\in X$ such that
$\rho(x,x_0)<\delta$} $$
In other words, $f(x)\in N_\sigma(f(x_0,\eps)\subseteq U$ for all $x\in
N_\rho(x_0, \delta)$. This says that whenever $x\in N_\rho(x_0, \delta)$
then $f(x)\in U$, and so $x\in f^{-1}(U)$. Thus,
$$N_\rho(x_0, \delta)\subseteq f^{-1}(U)$$
which shows that $f^{-1}(U)$ is open.
\vskip5pt
#end
#tex
\noindent
Now suppose that $f^{-1}(U)$ is open for every open set $U$, and show
that $f$ is continuous. To do this, we must show that $f$ is continuous
at every point in $X$. Let $x_0\in X$ and suppose $\eps>0$ is given. By
Proposition~5.13, the set $N_\sigma(f(x_0),\eps)$ is open, and so
$$f^{-1}\left( N_\sigma(f(x_0),\eps) \right) $$
is open too. Since $f(x_0)\in N_\sigma(f(x_0),\eps)$, this says that $x_0\in
f^{-1}\left( N_\sigma(f(x_0),\eps) \right)$. Since this is an open set,
there is a $\delta>0$ such that
$$N_\rho(x_0, \delta)\subseteq f^{-1}\left( N_\sigma(f(x_0),\eps) \right)$$
This last set inclusion says that:
\begin{quote}
\it
If $\rho(x,x_0)<\delta$ then $x\in N_\rho(x_0,\delta)$ and so $x\in
f^{-1}\left( N_\sigma(f(x_0),\eps) \right)$. This says that $f(x)\in
N_\sigma(f(x_0),\eps)$ and so $\sigma(f(x), f(x_0))<\eps$.
\end{quote}
This is exactly what we need to conclude that $f$ is continuous at $x_0$.
#end
#end
#ref(line)
#ref(remark)
Note that the preimage of an open set under a continuous function is open, but that the image of an open set need not be. The next example illustrates this.
#ref(line)
#ref(example)
Again, let
#ref(ex11)
Then:
#tex(ex16)
The image of an open set need not be open:
$$f(\;(-1,1)\;)=[0,1)$$
#end
#ref(ex16)
#link(Ex16)#ref(question)#end
Prove it.
#sideline(Ex16)
#ref(ex16)
#ref(line)
#tex
\begin{eqnarray*}
y\in f(\;(-1,1)\;) & \qquad\Longleftrightarrow \qquad & y=f(x)=x^2\mbox{ for some $x\in(-1,1)$} \\
& \qquad\Longleftrightarrow \qquad & y=x^2 \mbox{ for some $-1
#ref(line)
#tex(ex17)
However, the image of some open sets may be open, while the images of others is not:
$$f(\;(1,2)\;)=(1,4)$$
#end
#ref(ex17)
#sideline(Ex17)
#ref(ex17)
#ref(line)
#tex
\begin{eqnarray*}
y\in f(\;(1,2)\;) & \qquad\Longleftrightarrow \qquad & y=f(x)=x^2\mbox{ for some $x\in(1,2)$} \\
& \qquad\Longleftrightarrow \qquad & y=x^2 \mbox{ for some $1
#ref(line)
#tex(cor515)
{\bf Corollary 5.15.}
The map $f:X\rightarrow Y$ is continuous if and only if $f^{-1}(K)$ is closed for every closed set $K$ in $Y$.
#end
#ref(cor515)
#ref(line)
The proof of the corollary is based on Proposition 5.12, coupled with the following fact about preimages:
#ref(factoid)
#tex(fact1)
If $f:X\rightarrow Y$ and $B\subseteq Y$ then
$$f^{-1}(B^c)=(f^{-1}(B))^c$$
#end
#ref(fact1)
#link(Fact1)Proof.#end
#sideline(Fact1)
#ref(purple)Factoid:
#ref(fact1)
#ref(line)
#tex
\begin{eqnarray*}
x\in f^{-1}(B^c) & \qquad\Longleftrightarrow \qquad & f(x)\in B^c\\
& \qquad\Longleftrightarrow \qquad & f(x)\not\in B
\end{eqnarray*}
But $f(x)\in B$ if and only if $x\in f^{-1}(B)$ so this last statement is equivalent to
$$x\not\in f^{-1}(B) \qquad\Longleftrightarrow \qquad x\in (f^{-1}(B))^c$$
#end
#end
#ref(line)
#ref(remark)
By just the same token as before, the images of closed sets need not be closed. The images of some closed sets might be closed, but in general, continuous maps do not map closed sets to closed sets. (For example f(x)=1/(1+x^2) is continuous on the whole real line, but maps the set of natural numbers to a set which has 1 as a limit point, but does not contain it.)
There is however one important case in which closed sets are mapped to closed sets. All closed and bounded subsets of the real line share a property called compactness. The next chapter will be devoted to the study of compactness, and one of the things we'll see is that continuous functions always map compact sets to compact sets, and that compact sets are always closed (and bounded). Thus continuous functions will always map closed bounded sets in the real line to closed sets.