# The Unapologetic Mathematician

## 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 $d_Y(f(x),f(y))<\epsilon$.

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Posted by John Armstrong | Topology

## 16 Comments	»

1. 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

2. 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 | December 11, 2007 | Reply

3. [...] to verify this by a number of methods. Try using the pythagorean distance formula to make this a metric space, or you could work out a subbase of the product topology. In fact, not only should you get the same [...]

Pingback by | January 3, 2008 | Reply

4. [...] over at The n-Category Café reminded me of an interesting fact I haven’t mentioned yet: a metric space is actually an example of an enriched [...]

Pingback by | February 11, 2008 | Reply

5. [...] if for every there is a so that implies . But this talk of and is all designed to stand in for neighborhoods in a metric space. Picking a defines a neighborhood of the point . All we need is to come up with a notion of a [...]

Pingback by | April 19, 2008 | Reply

6. [...] know about products of topological spaces. We can take products of metric spaces, too, and one method comes down to us all the way from [...]

Pingback by | August 19, 2008 | Reply

7. [...] give the answer to the problem of pointwise convergence. It’s analogous to the notion of uniform continuity in a metric space. In that case we noted that things became nicer if we could choose our the same for every point, [...]

Pingback by | September 5, 2008 | Reply

8. [...] vector which points from one to the other. But a notion of distance is captured in the idea of a metric! So whatever a norm is, it should give rise to a metric by defining the distance as the norm of [...]

Pingback by | April 21, 2009 | Reply

9. [...] There’s an interesting little identity that holds for norms — translation-invariant metrics on vector spaces over or — that come from inner products. Even more interestingly, it [...]

Pingback by | April 24, 2009 | Reply

10. [...] it turns out that is a metric space, so all of the special things we know about metric spaces can come into play. Indeed, inner [...]

Pingback by | September 15, 2009 | Reply

11. [...] real spaces in hand, we can discuss continuous functions between them. Since these are metric spaces we have our usual definition with and and all that: A function is continuous at if and only if [...]

Pingback by | September 16, 2009 | Reply

12. [...] provides a nice way to restate our condition for continuity, and it works either using the metric space definition or the neighborhood definition of continuity. I’ll work it out in the latter case for [...]

Pingback by | December 7, 2009 | Reply

13. [...] as Metric It turns out that a measure turns its domain into a sort of metric space, measuring the “distance” between two sets. So, let’s say is a measure on an [...]

Pingback by | March 24, 2010 | Reply

14. [...] if we define a norm we get a “normed vector space”. This is a metric space, with a metric function defined by . This is nice because metric spaces are first-countable, and [...]

Pingback by | May 12, 2010 | Reply

15. [...] Metric Space of a Measure Ring Let be a measure ring. We’ve seen how we can get a metric space from a measure, and the same is true here. In fact, since we’ve required that be positive [...]

Pingback by | August 6, 2010 | Reply

16. [...] The metric only tells us about the lengths of tangent vectors; it is not a metric in the sense of metric spaces. However, if two curves cross at a point we can use their tangent vectors to define the angle [...]

Pingback by | September 20, 2011 | Reply

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