Metrics

November 26, 2010, 12:14
Filed under: Analysis, Math, Topology | Tags: analysis, Math, metric topology, pretty pictures, topology

It looks like, in the homestretch, I’ve been unable to post every day, so I’m going to consider myself out of MaBloWriMo. Which is a pity. On the other hand, I’d prefer, in the end, to post better posts less frequently and on a wider variety of topics.

In the interests of total confusion, let’s discuss metrics, with have almost nothing to do with measures, despite the similarity in name! Measures have to do with the sizes of sets of points, and are defined only on $\sigma$ -algebras. Metrics have to do with the distance between points, and are everywhere defined. Metric topology was perhaps the earliest field of topology to be studied, and so it’s not surprising that a metric on a space will give you a natural topology. Going the other way, from topologies to metrics, was a central problem of point-set topology in the 20th century.

A metric on a set $X$ (called a metric space) is a function $d:X\times X\rightarrow X$ with the following properties:

Symmetry: $d(x,y)=d(y,x)$ for all $x,y\in X$ .
Non-negativity: $d(x,y)\ge 0$ for all $x,y$ , and it equals $0$ iff $x=y$ .
Triangle inequality: $d(x,z)\le d(x,y)+d(y,z)$ for all $x,y,z\in X$ .

Just like we created measures from intuitive ideas about how area should work, all of these properties express intuitive ideas about distance. The third one is called the “triangle inequality” because it is an abstraction of the well-known fact that the sum of the lengths of two sides of a triangle is greater than the length of the third side.

We already know at least one metric in $\mathbb{R}^n$ : $d((x_1,\dotsc,x_n),(y_1,\dotsc,y_n)=\sqrt{(x_1-y_1)^2+\dotsb+(x_n-y_n)^2}$ . I guess we can call this the Euclidean metric. More generally, for any normed vector space, the norm induces a metric: $d(x,y)=\|x-y\|$ . This metric is then addition-invariant ( $d(x+a,y+a)=d(x,y)$ ) and satisfies $d(cx,cy)=cd(x,y)$ for $c$ a scalar.

Another good name for the Euclidean metric would be the 2 metric, because for any $1\le p<\infty$ , $\sqrt[p]{(x_1-y_1)^p+\dotsb+(x_n-y_n)^p} also defines a metric, the p metric. For $p=1$ , this is the Manhattan metric, defined by $d_1(x,y)=|x_1-y_1|+\dotsb+|x_n-y_n|$ . We can also take the limit as $p\rightarrow\infty$ and define the $\infty$ metric or sup metric by $d_{\infty}(x,y)=\sup |x_i-y_i|$ .¹

Given a metric, we can define the circle around $x$ of radius $\epsilon$ as $\{y:d(x,y)=\epsilon\}$ , and the ball of the same radius as $B_\epsilon(x)\{y:d(x,y)\le\epsilon\}$ . Circles in $\mathbb{R}^2$ in the 1,2, and $\infty$ metrics are to the right.

Now, the standard topology on $\mathbb{R}^n$ was given by a basis of Euclidean balls. Similarly, any metric space has an induced metric topology, given by the basis of balls $\{B_\epsilon(x):x\in X,\epsilon>0\}$ . If a metric induces a given topology on a set, we say that topological space is metrizable.

Let’s look at the p metrics on $\mathbb{R}^n$ , for example. Any ball in the 2-metric of radius $\epsilon$ contains the 1-metric ball of the same radius, and is contained by the 1-metric ball of radius $\sqrt{n}\epsilon$ . In particular, given any point and any open 2-metric ball, we can find a 1-metric ball containing that point and inside that 2-metric ball, and vice versa. Not sure if I’ve ever explicitly said this, but it’s pretty easy to prove that one topology is finer than another iff given a point and a basis element containing it in the second topology, you can always find a basis element of the first topology containing the point and inside the second basis element. So both topologies are finer than each other, so they’re equal. The same goes for any p, including infinity.

With metrics, unlike with topologies, you can define bounded sets. A bounded set is just a subset of some ball. But we showed that boundedness wasn’t preserved under homeomorphism, and likewise, for any metric, you can find a metric which gives the same topology but in which every set is bounded. Simply set $d^\prime(x,y)=\max\{d(x,y),1\}$ ; we can call this the bounded metric corresponding to $d$ . Balls of radius smaller than 1 are the same, but all balls of radius 1 or above equal the whole space, and the same argument as above shows that the topologies are equal. It’s a quirk of topology that it’s controlled by “small” sets in this way.

Subspaces of metric spaces inherit the metric, and yes, this induces the subspace topology (unlike the inherited order in a totally ordered set). For a finite product of metric spaces, you can set $d_\times(x,y)=d_1(p_1(x),p_1(y))+\dotsb+d_n(p_n(x),p_n(y))$ ; likewise, for a countable product, you can use $\sum_{i=1}^\infty\frac{d_i(p_i(x),p_i(y))}{2^i(1+d_i(p_i(x),p_i(y)))}$ . Both these induce the product topology. Uncountable products don’t, in general, have a metric, and we’ll see why below. For disjoint unions, I guess you could take the bounded metric on each component and set $d(x,y)=1$ if $x,y$ are in different components.

Munkres really likes putting metrics on $\mathbb{R}^\omega$ , but that doesn’t seem to be as important as “standard” analytical metric space properties, so I’ll skip it for now. Maybe it’ll be worth a quick post later.

Basically, all the analytical properties we generalized in topology come back to us in metric spaces. For example, given $f:X\rightarrow Y$ , where $X$ and $Y$ are metric spaces, $f$ is continuous at $x\in X$ iff for all $y\in X,\epsilon>0$ , there exists a $\delta>0$ such that if $d_X(x,y)<\delta$ , then $d_Y(f(x),f(y))<\epsilon$ . This is easy to prove using balls and the definition of continuity. Another statement shows us that sequences work well:

Lemma (the sequence lemma). Let $X,Y$ be topological spaces, $x\in X$ . If there is a sequence $(x_n)$ in $A\subset X$ that converges to $x$ , then $x\in\overline{A}$ . Likewise, if $f:X\rightarrow Y$ is continuous, then for every sequence $(x_n)$ converging to $x$ , $f(x_n)$ converges to $f(x)$ . The converses hold if $X$ is metrizable.

Proof. If $x_n\rightarrow x$ for $x_n\in A$ , then every neighborhood of $x$ contains a point in the sequence, which is a point of $A$ . So $x$ is in the closure of $A$ . Likewise, since continuous functions send neighborhoods to neighborhoods, given any neighborhood $B$ of $f(x)$ , we get a neighborhood $f^{-1}(B)$ of $x$ , that then must contain some $x_n$ . Then $f(x_n)\in B$ .

For the other direction, let $X$ have the metric $d$ . If $x\in\overline{A}$ , then each ball $B_{1/n}(x)$ intersects $A$ , so choose $x_n$ to be a point in its intersection. Clearly this sequence converges to $x$ . The last part — the converse of the second statement — is an exercise. $\square$

Often, we can use this to show that a space is not metrizable. For example, look at $\omega_1+1$ , where $\omega_1$ is the first uncountable ordinal, in the order topology. As we showed, no sequence of countable ordinals can converge to an uncountable one, so even though the final element $\omega_1$ is in the closure of the rest of the set, there isn’t a sequence converging to it. So this set isn’t metrizable. Similarly, look at an uncountable product like $\mathbb{R}^\mathbb{R}$ in the product topology. Let $A$ be the set of tuples $(x_i)$ which are $1$ for all but finitely many $i$ . The tuple which is all zeros is in the closure of this set: any neighborhood of it is $\mathbb{R}$ in all but finitely many coordinates, so it contains the tuple in $A$ which is $1$ in those all-but-finite coordinates and $0$ in the (finitely many) other ones. But any sequence of points in $A$ converges to a point which is $1$ in all but countably many coordinates! So $\mathbb{R}^\mathbb{R}$ is not metrizable.

From here, it’s worth taking a look at uniform continuity and uniform convergence, and then topological spaces whose elements are functions. In order to remedy our problem with sequences in sets like the above, we need to introduce nets, which are like sequences indexed on larger sets than $\mathbb{N}$ . As a kind of goofy thing, I can take a look at structures in between topological spaces and metric spaces, called uniform spaces, where you can define neighborhoods “of the same size” around any two points, though you can’t actually measure that size. (I don’t know that much about these guys, but I love to learn for this blog.) Finally, the metrization problem asks which topological spaces are metrizable and was a major topic of the mid-20th century. To answer it, we’ll have to delve into countability and separation axioms.

¹“Supremum” (“sup”) and “infimum” (“inf”) are very useful math words, by the way. They mean, respectively, “greatest lower bound” and “least upper bound.” For a finite number of things, as above, they just mean “maximum” and “minimum.” But an infinite number of things could approach an asymptote, in which case the max/min wouldn’t exist, but the sup/inf would.

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