The Proof is Trivial!

Though can you explain this step; not sure I quite follow (in particular the second term on the last line):

I did this:

Right. I was wondering if "cimer" was what people did when they got to the top of a mountain or something. Are you bilingual?

...

Ah I remember being fascinated by these numbers

Problem 585: If $S$ is countably infinite, show there exists an uncountable collection $\{A_i\}_{i \in I}$ of subsets of $S$ such that each $A_i$ is infinite and $A_i \cap A_j$ is at most finite for $i \neq j$.

Found this in a functional analysis text, though the proof is elementary. As an aside, it can be used to show that there does not exist a bounded projection $P : \ell_{\infty} \rightarrow c_0$, which in turn implies that $c_0$ is not isomorphic to a dual space.

Biject to the rationals via countability, so it is enough to consider $\mathbb{Q}$ (just invert/transport everything back along the bijection).

For every $\alpha \in \mathbb{R}$ pick a sequence of rationals $A_{\alpha} = \{\alpha_n \}_{n\in \mathbb{N}}$ converging to $\alpha$. For two sequences $\alpha_n$ and $\beta_n$, they would define a common subsequence and so $\alpha = \beta$. So we get an uncountable collection $\{A_{\alpha}\}_{\alpha \in \mathbb{R}}$ with pairwise finite intersections.

Don't know enough functional analysis to understand your last paragraph, but it sounds cool: $\ell_{\infty}$ is set of (bounded?) sequences equipped with the sup norm and $c_0$ is the set of sequences converging to $0$, right? What's a bounded projection and why does the above imply one doesn't exist? What's a dual space? Why does the above imply $c_0$ is not isomorphic to a dual space? Initial thought was that all dual spaces were isomorphic to $\ell_{0}$ but that doesn't seem anywhere near correct just via cardinality considerations.

You're correct about the definitions of the spaces (and yes we only consider bounded sequences). The notation is a bit sloppy, but viewing $c_0 \subset \ell_{\infty}$, a projection in this context is a bounded linear map $P : \ell_{\infty} \rightarrow \ell_{\infty}$ such that $P^2=P$ and $P(\ell_{\infty}) = c_0$. Also in the context of functional analysis we require all maps to be continuous, so it's much harder for spaces to be isomorphic unlike in linear algebra (we need a linear homeomorphism). In the same spirit, if $X$ is a normed space then its dual $X^*$ is the set of continuous linear functionals $P : X \rightarrow \mathbb R$ or $\mathbb C$.

Proving this claim takes a bit more work, but the idea is to apply the result to $S = \mathbb N$ and show that if $f \in (\ell_{\infty})^{\ast}$ then the set $\{i \in I : f(\chi_{A_i}) \neq 0 \}$ is countable ($\chi_{A_i}$ is the indicator function/sequence of each $A_i$). If such a projection $P$ exists, then $c_0 = \mathrm{Ker}\, (\mathrm{id}-P) = \bigcap_{n \in \mathbb N} e_n^{\ast} \circ (\mathrm{id}-P)$, where each $e_n^{\ast} \in (\ell_{\infty})^{\ast}$ is the nth coordinate functional. But this is an intersection of linear functionals, so by cardinality considerations it contains some $\chi_{A_i}$, which is a contradiction as each $\chi_{A_i} \not\in c_0$.

The last claim follows from a general fact that if $X$ is a normed space, there exists a bounded projection $P : X^{***} \rightarrow X^*$. The idea is to take the canonical embedding $i : X \rightarrow X^{**}$ and show its adjoint is a projection making appropriate identifications. We can apply this result here since $c_0^{**} \cong \ell_{\infty}$, making sure everything still works up to isomorphism.

Thanks for that; I understood the general gist of it. Functional Analysis seems really cool, definitely something I'll take next year. What book are you using, by the way?

Problem 588*

Put finitely many points in the plane, scattered such that they are not all in a line. Show that there is a line going through exactly two of them.

I've seen the proof of this before, so won't answer it here. But it really is gorgeous (especially considering when you take into account when it was proved).

Problem 591**
Here's one I came across recently. Evaluate $\displaystyle \int^\infty_0\frac{x}{e^x-1}\ dx$

Problem 591**
Here's one I came across recently. Evaluate $\displaystyle \int^\infty_0\frac{x}{e^x-1}\ dx$.

A hint for A Level students:

Where did you stumble upon this, out of interest? I also found it in a book that I was reading. There's actually a whole thread dedicated to tricky integrals, if you're not already aware

Thank you for the link. I got this question from someone I know who's doing maths at university.

You're correct about the definitions of the spaces (and yes we only consider bounded sequences). The notation is a bit sloppy, but viewing $c_0 \subset \ell_{\infty}$, a projection in this context is a bounded linear map $P : \ell_{\infty} \rightarrow \ell_{\infty}$ such that $P^2=P$ and $P(\ell_{\infty}) = c_0$. Also in the context of functional analysis we require all maps to be continuous, so it's much harder for spaces to be isomorphic unlike in linear algebra (we need a linear homeomorphism). In the same spirit, if $X$ is a normed space then its dual $X^*$ is the set of continuous linear functionals $P : X \rightarrow \mathbb R$ or $\mathbb C$.

Proving this claim takes a bit more work, but the idea is to apply the result to $S = \mathbb N$ and show that if $f \in (\ell_{\infty})^{\ast}$ then the set $\{i \in I : f(\chi_{A_i}) \neq 0 \}$ is countable ($\chi_{A_i}$ is the indicator function/sequence of each $A_i$). If such a projection $P$ exists, then $c_0 = \mathrm{Ker}\, (\mathrm{id}-P) = \bigcap_{n \in \mathbb N} e_n^{\ast} \circ (\mathrm{id}-P)$, where each $e_n^{\ast} \in (\ell_{\infty})^{\ast}$ is the nth coordinate functional. But this is an intersection of linear functionals, so by cardinality considerations it contains some $\chi_{A_i}$, which is a contradiction as each $\chi_{A_i} \not\in c_0$.

The last claim follows from a general fact that if $X$ is a normed space, there exists a bounded projection $P : X^{***} \rightarrow X^*$. The idea is to take the canonical embedding $i : X \rightarrow X^{**}$ and show its adjoint is a projection making appropriate identifications. We can apply this result here since $c_0^{**} \cong \ell_{\infty}$, making sure everything still works up to isomorphism.