The Unapologetic Mathematician

Mathematics for the interested outsider

We need to get down a few facts about metric spaces before we can continue on our course. Firstly, as I alluded in an earlier comment, compact metric spaces are sequentially compact — every sequence has a convergent subsequence.

To see this fact, we’ll use the fact that compact spaces are the next best thing to finite. Specifically, in a finite set any infinite sequence would have to hit one point infinitely often. Here instead, we’ll have an accumulation point $\xi$ in our compact metric space $X$ so that for any $\epsilon>0$ and point $x_m$ in our sequence there is some $n\geq m$ with $d_X(x_n,\xi)<\epsilon$. That is, though the sequence may move away from $\xi$, it always comes back within $\epsilon$ of it again. Once we have an accumulation point $\xi$, we can find a subsequence converging to $\xi$ just as we found a subnet converging to any accumulation point of a net.

Let’s take our sequence and define $F_N=\mathrm{Cl}(\{x_n, n\geq N\})$ — the closure of the sequence from $x_N$ onwards. Then these closed sets are nested $F_1\supseteq F_2\supseteq...\supseteq F_N\supseteq...$, and the intersection of any finite number of them is the smallest one, which is clearly nonempty since it contains a tail of the sequence. Then by the compactness of $X$ we see that the intersection of all the $F_N$ is again nonempty. Since the points in this intersection are in the closure of any tail of the sequence, they must be accumulation points.

Okay, that doesn’t quite work. See the comments for more details. Michael asks where I use the fact that we’re in a metric space, which was very astute. It turns out on reflection that I did use it, but it was hidden.

We can still say we’re looking for an accumulation point first and foremost, because if the sequence has an accumulation point there must be some subsequence converging to that point. Why not a subnet in general? Because metric spaces must be normal Hausdorff (using metric neighborhoods to separate
closed sets) and first-countable! And as long as we’re first-countable (or, weaker, “sequential”) we can find a sequence converging to any limit point of a net.

What I didn’t say before is that once we find an accumulation point there will be a subsequence converging to that point. My counterexample is compact, and any sequence in it has accumulation points, but we will only be able to find subnets of our sequence converging to them, not subsequences. Unless we add something to assure that our space is sequential, and metric spaces do that.

We should note in passing that the special case where $X$ is a compact subspace of $\mathbb{R}^n$ is referred to as the Bolzano-Weierstrass Theorem.

Next is the Heine-Cantor theorem, which says that any continuous function $f:M\rightarrow N$ from a compact metric space $M$ to any metric space $N$ is uniformly continuous. In particular, we can use the interval $\left[a,b\right]$ as our compact metric space $M$ and the real numbers $\mathbb{R}$ as our metric space $N$ to see that any continuous function on a closed interval is uniformly continuous.

So let’s assume that $f$ is continuous but not uniformly continuous. Then there is some $\epsilon>0$ so that for any $\delta>0$ there are points $x$ and $y$ in $M$ with $d_M(x,y)<\delta$ but $d_N(f(x),f(y))\geq\epsilon$. In particular, we can pick $\frac{1}{n}$ as our $\delta$ and get two sequences $x_n$ and $y_n$ with $d_M(x_n,y_n)<\frac{1}{n}$ but $d_N(f(x),f(y))\geq\epsilon$. By the above theorem we can find subsequences $x_{n_k}$ converging to $\bar{x}$ and $y_{n_k}$ converging to $\bar{y}$.

Now $d_X(x_{n_k},y_{n_k})<\frac{1}{n_k}$, which converges to ${0}$, and so $\bar{x}=\bar{y}$. Therefore we must have $d_Y(f(x_{n_k}),f(y_{n_k})$ also converging to ${0}$ by the continuity of $f$. But this can’t happen, since each of these distances must be at least $\epsilon$! Thus $f$ must have been uniformly continuous to begin with.

January 31, 2008 - Posted by | Point-Set Topology, Topology

1. So, what kinds of spaces are compact, but not sequentially compact? I don’t see how the fact that the space has a metric enters into the proof of the Bolzano-Weierstrass Theorem.

Comment by Michael Brazier | February 1, 2008 | Reply

2. Damn.. you’ve got a point there. All my references gave proof sketches for subspaces of $\mathbb{R}^n$ and then just asserted it worked for general metric spaces, so I tried to throw something together to work more generally, but it doesn’t seem to work. Thanks for catching that.

As for an example where compactness doesn’t imply sequential compactness, the best one I know of is the space of all functions from the unit interval to the set $\{0,1\}$, using the topology of pointwise convergence which I haven’t really discussed yet, but which is equivalent to the product topology on a continuum’s worth of copies of $\{0,1\}$. This is compact by the Tychonoff theorem, but a diagonalization argument can show that there’s no convergent subsequence.

Comment by John Armstrong | February 1, 2008 | Reply

3. Can you give a more or less explicit example of such a s sequence?

Comment by Michael Livshits | February 6, 2008 | Reply

4. In answer to Michael Livshits, I’d be fairly surprised if the following does not work (tweaking John’s example just a bit): consider all functions f: [0, 1] –> [0, 1] with the topology of pointwise convergence, and define a sequence f_n where f_n takes a real number r in [0, 1] to the fractional part of nr. If this had a convergent subsequence, indexed by an increasing set of integers a_n, then it would mean that the fractional parts {a_n r} would converge for every r in [0, 1]. My intuition says, to the contrary, that no matter what increasing sequence a_n you pick, there should be an r such that {a_n r} is equidistributed.

If that intuition is correct, then you can apply similar reasoning to John’s example (round down to 0 if the fractional part is less than 1/2; otherwise round up to 1).

Comment by Todd Trimble | February 6, 2008 | Reply

5. On further thought, the idea behind my example becomes even clearer if you consider instead the sequence of functions f_n(r) = {10^n r}. For any increasing sequence of natural numbers a_n, you can find an explicit r whose decimal expansion has, in its (a_{n}+1)th place after the decimal point, the digit given by the remainder of n mod 10 (take all the other digits of r to be 0, say). Quite clearly, the sequence of fractional parts {10^{a_n} r} cannot converge for that r. Neither does the sequence of 0’s and 1’s gotten by rounding up to 1 or rounding down to 0.

Comment by Todd Trimble | February 6, 2008 | Reply

6. Nice example, Todd. Replacing $n$ by $n_k$ in your argument, we can see that no subsequence of your sequence convereges pointwise.

Comment by Michael Livshits | February 6, 2008 | Reply

7. It looks like you can use $f_n(r)=sin(2^nr)$ instead.

Comment by Michael Livshits | February 6, 2008 | Reply

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