Metric Spaces and continuity of real-valued functions

Now that we’ve got the real numbers which correspond to our usual notion of magnitudes like distances, let’s refine our concept of a uniform space to take into account this idea of distance.

A metric space is a set equipped with a notion of “distance” in the set. That is, we have a set $M$ and a function $d:M\times M\rightarrow\mathbb{R}$ , which assigns a real number to every pair of points in $M$ . This function will satisfy the following axioms:

For all $x,y\in M$ , $d(x,y)\geq0$ .
For all $x,y\in M$ , $d(x,y)=0$ if and only if $x=y$ .
For all $x,y\in M$ , $d(x,y)=d(y,x)$ .
For all $x,y,z\in M$ , $d(x,z)\leq d(x,y)+d(y,z)$ .

The first says that distances are all nonnegative real numbers. The second says that any point is distance ${0}$ from itself, and only from itself. The third says that the distance between two points doesn’t depend on the order in which we take the points. The last is called the triangle inequality, because if we think of the points as the vertices of a triangle then it’s shorter to go from $x$ to $z$ along the leg connecting those two than to take the detour to $y$ .

Notice that these properties line up with those of absolute values. That is, the function $d:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ defined by $d(x,y)=\left|x-y\right|$ will be a distance function on $\mathbb{R}$ .

Now any metric space is actually a uniform space. We define an entourage $E_\delta$ for each positive real number $\delta>0$ . This $E_\delta$ consists of all the pairs $(x,y)\in M\times M$ with $d(x,y)<\delta$ . Since $d(x,x)=0$ each of these will contain the diagonal. The intersection $E_\delta\cap E_{\delta'}$ is the entourage corresponding to the smaller of $\delta$ and $\delta'$ . Each $E_\delta$ is its own reflection by the symmetry of the distance function. And the triangle inequality gives our half-size entourages — if $(x,y)\in E_{\frac{\delta}{2}}$ and $(y,z)\in E_{\frac{\delta}{2}}$ then $(x,z)\in E_\delta$ .

For the real numbers themselves we should verify that we get back the same uniform structure as we did before. Remember that the uniform structure we got from completing the uniform structure on the rational numbers had an entourage $E_r$ for each positive $r\in\mathbb{Q}$ , with $E_r=\{(x,y)\in\mathbb{R}\times\mathbb{R}|\left|x-y\right|^lt;r$ . Each one of these shows up in the entourages for the metric structure, by considering $r$ as a positive real number, but does every basic entourage from the metric structure show up as an entourage in the complete uniform structure? It does! The Archimedean property tells us that for any positive $\delta$ we can find a positive rational number $r<\delta$ . Then $E_r\subseteq E_\delta$ , and so $E_\delta$ is an entourage in the completion of the uniform structure on the rationals.

Let’s look at the neighborhood structure we get from the entourages of the metric structure. A subset $U$ is a neighborhood of $x$ if and only if it contains $E_\delta\left[x\right]$ for some $\delta>0$ . That is, it must contain the “open ball” of all $y\in X$ such that $\left|x-y\right|<\delta$ .

In $\mathbb{R}$ this means that we have a neighborhood base for each point $x$ consisting of the intervals $(x-\delta,x+\delta)=\{y\in\mathbb{R}|\left|x-y\right|<\delta\}$ . Thus a subset $U$ of $\mathbb{R}$ will be open if and only if it contains such a symmetric neighborhood of each of its points, and this will happen if and only if $U$ is the union of a collection of open intervals. Then we can take the intervals $(a,b)=\{x\in\mathbb{R}|a<x<b\}$ as a base for our topology.

As a final coup de grâce, let’s write down explicitly the condition that a function $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous map. We have a neighborhood base of our topology, and we know we only need to check the neighborhood definition of continuity on a neighborhood base.

So, a function $f$ will be continuous at $x$ if and only if for each neighborhood $V\in\mathcal{N}(f(x))$ there is a neighborhood $U\in\mathcal{N}(x)$ with $f(U)\subseteq V$ . Translating this all into our explicit language for the real numbers and restricting to neighborhood bases says that a function $f$ is continuous at $x$ if and only if for each $\epsilon>0$ there is a $\delta>0$ so that $\left|x-y\right|<\delta$ implies $\left|f(x)-f(y)\right|<\epsilon$ . And we’re back to the old definition of continuity from calculus 1! Then, as usual, we say that $f$ is continuous if the above condition holds for all $x\in\mathbb{R}$ .

What about uniform continuity. We can again translate the statements to our special case and check them on the basic entourages. A function $f:\mathbb{R}\rightarrow\mathbb{R}$ will be uniformly continuous if for every $\epsilon>0$ there is a $\delta>0$ so that for all $x,y\in\mathbb{R}$ , $\left|x-y\right|<\delta$ implies that $\left|f(x)-f(y)\right|<\epsilon$ .

Notice particularly the difference between uniform continuity and continuity. Continuity says that (for all $x\in\mathbb{R}$ ) (for all $\epsilon>0$ there exists a $\delta>0$ ) such that ( $\left|x-y\right|<\delta$ implies $\left|f(x)-f(y)\right|<\epsilon$ ). Uniform continuity says that (for all $\epsilon>0$ there exists a $\delta>0$ ) such that (for all $x\in\mathbb{R}$ ) ( $\left|x-y\right|<\delta$ implies $\left|f(x)-f(y)\right|<\epsilon$ ). The quantifier for $x$ shows up after the quantifier for $\delta$ in the latter definition. That is, for a uniformly continuous function we can pick the $\delta$ uniformly to apply to all points $x$ , while for a merely continuous function we may have to use a different $\delta$ for each point $x$ . At first it doesn’t seem to be that big a deal, which always causes a certain amount of confusion in an advanced calculus (undergraduate real analysis) class, but it turns out that being able to choose the same $\delta$ at every point makes a lot of nice things work out that don’t otherwise hold.

[UPDATE]: I’m feeling a little silly that I didn’t mention this before, but the last two definitions immediately port over to any function between metric spaces $X$ and $Y$ by just using the local definitions of “distance” in place of that for $\mathbb{R}$ . A function $f:X\rightarrow Y$ is continuous if for all $x\in X$ and $\epsilon>0$ there is a $\delta>0$ so that $d_X(x,y)<\delta$ implies $d_Y(f(x),f(y))<\epsilon$ . Similarly, $f$ is uniformly continuous if for all $\epsilon>0$ there is a $\delta>0$ so that for all $x\in X$ $d_X(x,y)<\delta$ implies $d_Y(f(x),f(y))<\epsilon$ .

December 10, 2007 - Posted by John Armstrong | Topology

16 Comments »

John, could you correct some typos in UPDATE (and also in the text). Because the variable y looks like as a free variable in definitions of the (uniform) continuity.

—
Z. Karno

Comment by Zbigniew Karno | December 11, 2007 | Reply
There’s no typo there. The quantifier on $y$ is implicit, just as it is for the definition of continuity, and both definitions in the case of $\mathbb{R}$ . I’m not writing well-formed formulæ in a predicate calculus here, and it’s common to informally let a variable range over whatever values make sense. I’m sure I’ve done it a dozen other places before this.

Comment by John Armstrong | December 11, 2007 | Reply
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