Topology | 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.

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

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October 26, 2008

Random Variables

Filed under: Measure Theory,Probability Theory,Topology — cjohnson @ 8:32 pm

If $(\Omega, \mathcal{F}, P)$ is a probability space and $(Y, \Sigma)$ is a measurable space (i.e., a set $Y$ along with a sigma-algebra $\Sigma$ on $Y$ ), then a random variable is a measurable function $X : \Omega \to Y$ . That is, for each $A \in \Sigma$ , we have $X^{-1}(A) = \{ \omega \in \Omega : X(\omega) \in A \} \in \mathcal{F}$ . Generally speaking, we’ll be taking $(Y, \Sigma)$ to be $(\mathbb{R}, \mathcal{B})$ where $\mathcal{B}$ is the Borel algebra on $\mathbb{R}$ .

Given a topological space $(S, \tau)$ , there exists a sigma-algebra $\mathcal{B}(S)$ , called the Borel algebra on $S$ , that contains all open sets (members of $\tau$ ) and is the smallest such sigma-algebra. This means that for each $\mathcal{O} \in \tau$ we have $\mathcal{O} \in \mathcal{B}(S)$ and also if $\Sigma$ is another sigma-algebra on $S$ with this property, then $\mathcal{B}(S) \subseteq \Sigma$ . In general, given a collection $C$ of subsets of $S$ , there exists a sigma-algebra, which we’ll call $\mathcal{F}_C$ , that is the smallest sigma-algebra on $S$ (smallest using $\subseteq$ as the ordering relation) containing each $c \in C$ ; we say that $\mathcal{F}_C$ is the sigma-algebra generated by $C$ . In this sense, the Borel algebra on $S$ is the sigma-algebra generated by the topology $\tau$ . We will usually just write $\mathcal{B}$ to mean $\mathcal{B}(\mathbb{R})$ .

In the special case of $(\mathbb{R}, \mathcal{B})$ , there’s an easy way to check to see that a function $X : \Omega \to \mathbb{R}$ is a random variable (or a measurable function in general): we just look at the pre-images of intervals. Since the pre-images of functions are well-behaved with respect to set operations like union, intersection, and complement, it in fact suffices to only consider pre-images of the form ${}[a, \infty)$ . That is, if we show that $X^{-1}([a, \infty)) \in \mathcal{F}$ for every $a \in \mathbb{R}$ , we will have shown that $X$ is measurable, and so a random variable. (Actually, we can look at all intervals of the form $(a, \infty)$ , ${}[a, \infty)$ , $(-\infty, a)$ or $(-\infty, a]$ . Using properties of sigma-algebras we can easily show that if we have all intervals of any of these forms, we have all intervals of any other form. Again, using properties of sigma-algebras it’s easy to take that and show that we have all countable unions of intervals — namely all countable unions of open intervals, i.e., all open sets.)

To see this, suppose for every $a \in \mathbb{R}$ we have $X^{-1}([a, \infty)) \in \mathcal{F}$ .

$\displaystyle X^{-1}([a, \infty)) \in \mathcal{F}$

$\displaystyle \implies \{ \omega \in \Omega : X(\omega) \geq a \} \in \mathcal{F}$

$\displaystyle \implies \{ \omega \in \Omega : X(\omega) \geq a \}^\complement \in \mathcal{F}$

$\displaystyle \implies \{ \omega \in \Omega : X(\omega) < a \} \in \mathcal{F}$

$\displaystyle \implies X^{-1}((-\infty, a)) \in \mathcal{F}$

So now we have the pre-images of all intervals of the form ${}[a, \infty)$ and $(-\infty, a)$ . If we can also get pre-images for the form $(a, \infty)$ , it’ll be an easy jump to countable unions of open intervals. Note that

$\displaystyle (a, \infty) = \bigcup_{n=1}^\infty \left[ a + \frac{1}{n}, \infty \right)$

Now we easily see that

$\displaystyle X^{-1}((a, \infty)) = X^{-1} \left( \bigcup_{n=1}^\infty \left[ a + \frac{1}{n}, \infty \right)\right)$

$\displaystyle = \bigcup_{n=1}^\infty X^{-1}\left(\left[ a + \frac{1}{n}, \infty \right)\right)$

And we have pre-images of intervals of the form $(a, \infty)$ . Combined with the fact we have intervals of the form $(-\infty, b)$ , it’s easy to see that we also have intervals of the form $(a, b)$ . Using properties of sigma-algebras, it’s easy to show now that we have the pre-image of any open set. This tells us that if $X^{-1}([a, \infty)) \in \mathcal{F}$ for every $a \in \mathbb{R}$ , then we have that $X$ is measurable, and so a random variable.

In the case that $\Omega$ is a countable set, we take $\mathcal{F}$ to be $2^\Omega$ (the powerset of $\Omega$ ), and so any function $X : \Omega \to \mathbb{R}$ is a random variable. This is because the pre-image of ${}[a, \infty)$ must be something (even if it’s empty); we have for every $a \in \mathbb{R}$ that $X^{-1}([a, \infty)) \in 2^\Omega$ , and so every function on a countable sample-space is a random variable.

July 15, 2008

Some Basic Topology, Part I

Filed under: Analysis,Functional Analysis,Topology — cjohnson @ 12:16 pm

A topology on a set $X$ is a collection of subsets of $X$ , which we’ll call $\tau$ , such that the following are satisfied

$\emptyset, X \in \tau$
If $\Lambda$ is some index set where $\forall \lambda \in \Lambda : U_\lambda \in \tau$ , then $\bigcup_{\lambda \in \Lambda} U_\lambda \in \tau$
If $U_1, ..., U_n \in \tau$ for any $n \in \mathbb{N}$ , then $\bigcap_{i=1}^n U_i \in \tau$ .

If $(X, d)$ is a metric space, then we can generate a topology $\tau$ on $X$ by considering the collection of all “open balls” in $X$ .

Given any point $x$ in a metric space $(X, d)$ and an $r > 0$ , the open ball of radius $r$ centered at $x$ , denoted $B_r(x)$ , is the set of all points in the space that are less than distance $r$ from $x$ . That is, $B_r(x) = \{ y \in X : d(x, y) < r \}$ . The closed ball of radius $r$ centered at $x$ is defined similarly: $\overline{B_r}(x) = \{ y \in X : d(x, y) \leq r \}$ . Finally, the sphere of radius $r$ centered at $x$ consists of those points that are precisely distance $r$ from $x$ : $S_r(x) = \{ y \in X : d(x, y) = r \}$ .

If $M$ is some subset of $X$ and $m \in M$ , then we say that $m$ is an interior point of $M$ if there exists an $r > 0$ such that $B_r(m)$ is contained in $M$ . If every point of a set is an interior point, then we say that the set is open. For any set, open or not, we define the interior of the set to be all of the sets interior points.

Note that the interior of a set is the largest open set contained in that set. For example, suppose $M \subseteq X$ and let $M^\circ$ be the interior of $M$ (there isn’t, as far as I know, any real standard notation for the interior of a set). By the definition of the interior, $M^\circ \subseteq M$ and each point in $M^\circ$ can have an open ball placed around it that stays within the set, so $M^\circ$ is an open subset of $M$ . Suppose $I$ is also an open subset contained in $M$ and that $M^\circ \subset I \subseteq M$ . This means that there must exist an $x \in I$ and a corresponding $r > 0$ such that $B_r(x) \subseteq I \subseteq M$ . However, by assumption, $M^\circ$ is the interior of $M$ so it must contain all such points, and no such $I$ can exist, thus $M^\circ$ is the largest open subset contain in $M$

The point $x$ is called a limit point (some authors call it an accumulation point) of $M \subseteq X$ if for every $r > 0$ there exists a $m \in B_r(x)$ such that $m \in M$ and $x \neq m$ . Note that a limit point of a set need not be an element of that set. For instance, on the real line with the standard metric, 1 and 0 are both limit points of (0, 1) but neither is in the set. (As the name “limit point” implies, we can construct a sequence in $M$ whose limit is $x$ , but we’ll have to wait until we define the limit of a sequence later before we can show that.)

If $M \subseteq X$ and the complement of $M$ in $X$ , $M^c = X \setminus M$ , is open, then we say that $M$ is closed. The closure of a set $M$ , denoted $\overline{M}$ , is the set $M$ along with all of its limit points. This forms the smallest closed set that contains $M$ .

Note that a set (unlike a door … bad joke) may be open, closed, both, or neither. On the real line, for instance, the interval [0, 1) is neither open nor closed and the entire space, $\mathbb{R}$ is both open and closed. In any topology, the empty set, $\emptyset$ and the entire space, $X$ , are always both open and closed. They are both open since, by the definition of a topology, they must be an element in the topology. They are closed because they are complements of one another and the complement of an open set is a closed set. In the discrete topology (where each subset of the space is an element of the topology) all sets are both open and closed. (I’m not going to discuss it right now, but notice that the discrete topology is generated by the discrete metric. To show that we’d need to talk about the basis of a topology, which we’re not going to do right now.)

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