If you are like me, you’ve probably got lots to do for final exams and things of the like as the semester (trimester, quarter, whatever your school uses) comes to a close. In an effort to take a bit of break, I would like to present you with a cute little result I came across in my studies. It’s a nice little number theory fact with a cute combinatorial proof. I found the result in Hatcher’s book on algebraic topology, and I’ll present an adapted version of his proof (though I’m sure it’s in many other books).
First we need to remember a little about 
Proposition: Let 
where 
Proof: Recall that in the polynomial ring ![\mathbb{Z_p}[x]](https://s0.wp.com/latex.php?latex=%5Cmathbb%7BZ_p%7D%5Bx%5D&bg=ffffff&fg=333333&s=0&c=20201002)
On one hand, we know that the coefficient of 
And that’s it! I hope you enjoyed it, and now I’ve got to get back to work!
. This is because the first vote was for 
, but more importantly it means that since the beginning (when no one had a majority), 
, and after the third step it is 
. So, again, since the starting position when no one was winning, 
the count is at 
. Note: another way to look at the count at this point is that the number of 
, meaning at this point no candidate has a majority of the votes. After the next step, though, we are going to dictate that the the count be 
. Note that this pair does not mean that 
only means that since the last no-winner point, 
. I defined the topology in terms of net convergence, but remarked that this was equivalent to the weak topology generated by 
. The purpose of this post will be to introduce a bunch of different topologies we can put on a Banach space, its dual, and even the set of linear operators on the Banach space. Before I do this however, I want to spend a bit of time reviewing the notion of the weak topology on a space generated by a collection of functions.
be a topological space and 
is a collection of functions, then the weak topology generated by 
is the smallest topology on 
is continuous.
, where 
is open in the topology on 
be as in the above definition and endow 
in 
if and only if for every 
, the net 
converges to 
in 
, then by the continuity of 
(for all 
), we get the result. Conversely, suppose 
does not converge to 
. Since 
does not converge to 
containing 
, where 
is open (technically we should take distinct sets 
, but if we intersect these sets, the resulting smaller set works just fine). Now, since this is a finite intersection, the condition that 
. But this shows that 
is not eventually in 
, a neighborhood of 
, hence 
. Typically in point-set topology we are concerned with the first characterization of the product topology, namely it’s generated by finite intersections of cylinders 
, for 
. However, let’s see what happens if we consider this in terms of nets. For a finite product, then this says a net 
converges to 
if and only if 
and similarly 
. 
as the set of all functions from 
to 
with 
and 
? Well, this says that 
if and only if 
and 
, in other words it’s a type of pointwise convergence. Suppose 
denotes the continuous functions 
(or 
, either works). We can view 
, i.e. 
. If we give this space the product topology, then a net 
(which need not be in 
for each 
for all 
, since, for example, it’s easy to construct continuous functions on 
. However, a more natural topology to consider is the so called weak* topology. Recall that we can naturally embed 
, where 
, i.e. 
is the so-called evaluation functional. The weak* topology on 
. In terms of nets then, a net in 
, i.e. if and only if 
, the collection of bounded (i.e. continuous) operators 
. As usual, we have the norm topology (using the operator norm), and the topology of pointwise convergence, which is often called the strong operator topology. We can also consider the weak operator topology, which is best defined in terms of nets. We say 
converges to 
converges weakly to 
, so the weak operator topology might best be described as the topology of weak pointwise convergence. Now, without further wait, I give you (with proof adapted from Folland)
, i.e. to each 
and note that 
is compact by Tychonoff’s theorem. Now, 
on 
for all 
, all the remains to show is that 
and 
, we have
so that 
equipped with the supremum norm.
be the unit ball of 
in 
(it’s not immediately clear that 
, so indeed 
. Clearly the map 
is also bounded, hence 
is actually an isometric isomorphism onto its image). The last thing to prove is that 
is a net in 
. Then 
is a Cauchy net in 
, we have 
and the proof is complete.![[0,1]](https://s0.wp.com/latex.php?latex=%5B0%2C1%5D&bg=ffffff&fg=333333&s=0&c=20201002)
and use real valued functions instead of complex valued functions. That’s pretty cool! So, as you can see, the idea of weak topologies is useful and in some sense quite natural when considering normed spaces. Anyways, that’s about all I’ve got for this post, hopefully you enjoyed it.
-low polling numbers, you might suspect that most of the candidates must be working hard to lose the winnability property. Or maybe you thought that Sheldon Adelson would attempt to satisfy the dictatorial property. And yet, at least in theory, neither of those were it. It was the manipulability property that was actually being broken.
is a finite dimensional Hilbert space, then the unit ball of 
this is obviously true, since the unit ball is closed and bounded. Given any finite dimensional Hilbert space, it is isometrically isomorphic to 
. Under this isomorphism, the unit ball of 
are two norms on 
so that 
for all 
, and as an easy corollary to this we see that any two norms generate the same topology on 
, the set of square summable sequences of complex (or real) numbers with the natural inner product. Let 
be an orthonormal basis for 
, the easiest example is to take 
, where the 
occurs in the 
be a countable subset of the orthonormal basis. Now, for 
, we have
-separated sequence sitting in the unit ball, and so it has no convergent sub-sequence. For a metric spaces, sequential compactness and compactness are equivalent, so this shows the unit ball is not compact.
converges to 
, the net 
converges to 
implies 
converges to 
is given by 
for some fixed vector 
(and conversely, the map 
defines a bounded linear functional on 
weakly in 
, we have 
. Thus for Hilbert spaces, we can think of weak convergence as “convergence of inner products”. One might ask if this is actually a genuinely weaker topology than the norm topology on 
converges weakly to 
. Moreover, 
does not converge in the norm topology on 
. But this is precisely the statement that 
. It is left to the reader to verify that this implies 
as 
, i.e. 
weakly in 
and 
is countable, then there is a function that assigns each member 
of 
is not a countable union of countable sets. (Apparently the assertion the countable union of countable sets is countable is actually strictly weaker than 
th term from, e.g., a ball of radius 
. The proof of the 
is a relation on a nonempty set 
there is a 
such that 
, then there is a sequence 
, of members of 
for every 
. 
(this is less true of 
such that 
is a member of 
every set can be linearly ordered 
and 
then there is a bijection 
-coloring of the 
of 
, choose (by 
is countable. Contrast that statement and proof to the following:
are countably many sets, and for each 
is a bijection, then there is a bijection 
(and we can write down a formula for it in terms of the 
). 
from the 
would be: you can prove many things, but without understanding your subject as deeply as you would otherwise. (I should mention that I owe this example, related insight, and the expression ‘snorting choice’ to 
Adam Azzam
Jay Cummings
Ryan
Z Norwood
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