# April Is Separability Month

It is now April! Did you know that April is one of the months in which every compact metric space is separable?

Proof. Let it be April, and let M be a compact metric space. Because M is compact, it is totally bounded, so for all n∈ℕ, we can cover M with finitely many open balls of radius 1/n. The centers of all such balls are a countable set which we can call C. But C is dense, because an arbitrary point p∈M is a limit point of C: an ε-neighborhood of p must contain the center of one the balls in our covering of M with ε/2-balls. Thus M contains a countable dense subset.

# Lipschitz

—and the moment or more than a moment when the dam breaks, when the damned break and the void inside their skulls is filled (the atmosphere rushing in quickly, but not so quickly that one couldn't sense its motion) with the terror that is knowledge of the specter of continuity: that there have never been, and can never be, any miracles.

For to be saved is only to be some distance in the initial conditions from being damned, some lesser distance from being half-damned ... some δ-distance from being ε-damned. And the complement of the shadow we cast on the before-time contains its limits.

# Missing Books I

Someone should write a combined novel/textbook about a mathematician-princess's quest to understand the true nature of continuity and change. When her father dies, she'll have the opportunity to be Queen regnant, but she'll quickly marry some guy instead so she can be a Queen consort and continue her research without being distracted with boring politics.

# Two Views of the Monotone Sequence Theorem

If a sequence of real numbers (an) is bounded and monotone (and I'm actually going to say nondecreasing, without loss of generality), then it converges. I'm going to tell you why and I'm going to tell you twice.

If our sequence is bounded, the completeness of the reals ensures that it has a least upper bound, which we'll call, I don't know, B, but there have to be sequence elements arbitrarily close to (but not greater than) B, because if there weren't, then B couldn't be a least upper bound. So for whatever arbitrarily small ε, there's an N such that aN > B – ε, which implies that |aNB| < ε, but if the sequence is nondecreasing, we also have |anB| < ε for nN, which is what I've been trying to tell you—

twice; suppose by way of contraposition that our sequence is not convergent. Then there exists an ε such that for all N, there exist m and n greater or equal to N, such that |aman| is greater or equal to ε. Suppose it's monotone, without loss of generality nondecreasing; that implies that for all N, we can find n > mN such that anam ≥ ε. Now suppose our sequence is bounded above by some bound B. However, we can actually describe an algorithm to find sequence points greater than B, thus showing that this alleged bound is really not a bound at all. Start at a1. We can find points later in the sequence that are separated from each other by at least ε, but if we do this ⌈(Ba1)/ε⌉ times, then we'll have found a sequence point greater than the alleged bound.

# Subscripting as Function Composition

Dear reader, don't laugh: I had thought I already understood subsequences, but then it turned out that I was mistaken. I should have noticed the vague, unverbalized discomfort I felt about the subscripted-subscript notation, (ank). But really it shouldn't be confusing at all: as Bernd S. W. Schröder points out in his Mathematical Analysis: A Concise Introduction, it's just a function composition. If it helps (it helped me), say that (an) is mere syntactic sugar for a(n): ℕ → ℝ, a function from the naturals to the reals. And (ank) is just the composition a(n(k)), with n(k): ℕ → ℕ being a strictly increasing function from the naturals to the naturals.

# Bounded but Not Totally Bounded, Redux

Theorem. An open set in real sequence space under the ℓ norm is not totally bounded.

Proof. Consider an open set U containing a point p. Suppose by way of contradiction that U is totally bounded. Then for every ε > 0, there exists a finite ε-net for U. Fix ε, and let m be the number of points in our ε-net, which net we'll denote {Si}i∈{1, ..., m}. We're going to construct a very special point y, which does not live in U. For all i ∈ {1, ..., m}, we can choose the ith component yi such that the absolute value of its difference from the ith component of the ith point in the net is strictly greater than ε (that is, |yiSi,i| > ε) but also so that the absolute value of its difference from the ith component of p is less than or equal to ε (that is, |yipi| ≤ ε). Then for j > m, set yj = pj. Then |yp| ≤ ε, but that means there are points arbitrarily close to p which are not in U, which is an absurd thing to happen to a point in an open set! But that's what I've been trying to tell you this entire time.

# Bounded but Not Totally Bounded

The idea of total boundedness in metric space (for every ε, you can cover the set with a finite number of ε-balls; discussed previously on An Algorithmic Lucidity) is distinct from (and in fact, stronger than) the idea of mere boundedness (there's an upper bound for the distance between any two points in the set), but to an uneducated mind, it's not immediately clear why. What would be an example of a set that's bounded but not totally bounded? Wikipedia claims that the unit ball in infinite-dimensional Banach space will do. Eric Hayashi made this more explicit for me: consider sequence space under the ℓ norm, and the "standard basis" set (1, 0, 0 ...), (0, 1, 0, 0, ...), (0, 0, 1, 0, 0, ...). The distance between any two points in this set is one, so it's bounded, but an open 1-ball around any point doesn't contain any of the other points, so no finite number of open 1-balls will do, so it's not totally bounded, which is what I've been trying to tell you this entire time.