# Mathematics Prelims

## January 24, 2009

### Continuity in Topology

Filed under: Topology — cjohnson @ 3:20 pm

Recall a function $f : X \to Y$ between metric spaces $(X, d_1)$ and $(Y, d_2)$ is called continuous at a point $x_0 \in X$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $d(x, x_0) < \delta$ implies $d_2(f(x), f(x_0)) < \epsilon$.  That is, if $x$ is “near” $x_0$, then $f(x)$ will be “near” $f(x_0)$.  We will say $f$ is continuous if it’s continuous at every point in $X$.  We can extend continuity from metric spaces to general topological spaces by the following.

If $(X, \tau_1)$ and $(Y, \tau_2)$ are two topological spaces, then we say that a function $f : X \to Y$ is continuous if for every $U \in \tau_2$ we have $f^{-1}(U) \in \tau_1$; that is, if the preimage of every open set in $Y$ is an open set in $X$.  Before discussing this further, let’s recall some properties of preimages.

1. The preimage of a complement is the complement of the preimage: $f^{-1}(U^\complement) = f^{-1}(U)^\complement$.
2. The preimage of a union is the union of the preimages: $f^{-1}(U \cup V) = f^{-1}(U) \cup f^{-1}(V)$.
3. The preimage of an intersection is the intersection of the preimages: $f^{-1}(U \cap V) = f^{-1}(U) \cap f^{-1}(V)$
4. The preimage of a difference is the difference of the preimages: $f^{-1}(U \setminus V) = f^{-1}(U) \setminus f^{-1}(V)$

To see the first of these properties, simply write out the definitions for the preimages.

$\displaystyle f^{-1}(U^\complement) = \{ x \in X : f(x) \in U^\complement \}$

$\displaystyle = \{ x \in X : f(x) \notin U\}$

$\displaystyle = \{ x \in X : f(x) \in U \}^\complement$

$\displaystyle = f^{-1}(U)^\complement$

Showing the other equalities follows the same general format of writing out the definition for the preimage and following your nose.

Of course, to call this topological definition an extension of the definition for metric spaces, we must show that our definition for continuity in metric spaces is a special case of this topological definition.  Recall that a metric space is a topological space whose topology is generated by the collection of open balls; a subset of a metric space is open if around each point in the set we can place an open ball that stays entirely within the set.

Suppose $f : X \to Y$ is continuous in the metric sense.  We wish to show that $f$ is continuous in the topological sense as well.  Let $y \in Y$ and let $\epsilon > 0$ be given.  Note $B_\epsilon(y)$ is an open subset of $Y$.  Let $x \in f^{-1}(B_\epsilon(y))$.  This means $f(x) \in B_\epsilon(y)$.  We can then put a ball around $f(x)$ and stay inside of $B_\epsilon(y)$.  In particular, let $\eta = (\epsilon - d_2(f(x), y))/2$ and note $B_\eta(f(x)) \subseteq B_\epsilon(y)$.  Consider the preimage of $B_\eta(f(x))$.  By the assumption of continuity we can find a $\delta > 0$ such that $d_1(z, x) < \delta$ implies $d_2(f(z), f(x)) < \eta$.  The points satisfying $d_1(z, x) < \delta$ are precisely the points of $B_\delta(x)$ though.  All together this means if we find a point in the preimage of a ball, we can put a ball around that point that stays inside the preimage: the preimage of a ball is an open set.  If $U \subseteq Y$ is a general open subset of $Y$ (not necessarily a ball), then for each $u \in U$ let $\epsilon(u)$ denote a number such that $B_{\epsilon(u)}(u) \subseteq U$.  Then we have

$\displaystyle f^{-1}(U) = f^{-1} \left( \bigcup_{u \in U} B_{\epsilon(u)}(u) \right)$

$\displaystyle = \bigcup_{u \in U} f^{-1} \left(B_{\epsilon(u)}(u)\right)$

We’ve just showed the preimage of an open ball is an open set, so now we have that the preimage of an openset is a union of open sets, so it is an open set as well.

Now suppose $f : X \to Y$ is such that for every open $U \subset Y$ we have $f^{-1}(U)$ is open in $X$.  Then, in particular, for any $\epsilon > 0$  and $x \in X$ we have $f^{-1}(B_\epsilon(f(x)))$ is an open subset of $X$.  This open subset must contain $x$, so we can put a ball around $x$ staying inside the preimage.  Calling the radius of this ball $\delta$, we have that $f$ is continuous in the metric sense.

Defining continuity like this gives us a way to discuss continuous functions in spaces where we don’t necessarily have a metric, as long as we have a topology.  For instance, if $X = \{a, b, c\}$, then $\tau = \{ \emptyset, X, \{a\}, \{c\}, \{a,c\}\}$ is a topology on $X$.  We might define a function $f : \mathbb{R} \to X$ as

$\displaystyle f(x) = \left\{ \begin{array}{ll} a &: x < 0 \\ b &: 0 \leq x \leq 1 \\ c &: 1 < x \end{array} \right.$

Then $f$ would be a continuous function as $f^{-1}(\{a\}) = (-\infty, 0)$; $f^{-1}(\{c\}) = (1, \infty)$; $f^{-1}(\{a, c\}) = (-\infty, 0) \cup (1, \infty)$; and so on.

Even when we do have a metric space, it may be easier to show that the preimage of an open set is open, than it is to show that for every point and every $\epsilon > 0$ we can find a corresponding $\delta$.  These are equivalent, but it may be easier in some cases to deal with general open sets than with $\epsilon-\delta$-style proofs.