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I've been mulling this idea over for a while now with little progress. Maybe someone here will have more success with it.

Here, let $\ell_1$ be the set of real-valued sequences $\{x_n\}$ such that $\sum_{n\in\mathbb{N}}|x_n|\lt\infty$, and let $\ell_\infty$ be the set of real-valued sequences such that $\sup_{n\in\mathbb{N}}|y_n|\lt\infty$. Suppose I equip $\ell_\infty$ with the coarsest topology such that all functions mapping $\{y_n\}\mapsto\sum_{n\in\mathbb{N}}x_ny_n$, for $\{x_n\}\in\ell_1$, are continuous from $\ell_\infty$ to $\mathbb{R}$.

Is there a proof that the set $\{\{y_n\}\mid \sup_{n\in\mathbb{N}}|y_n|\leq 1\}$ is compact? Thanks.

Edit: Thank you for the responses so far. I wanted to add one thing. Is it possible to show this in a way using only techniques of general topology? I haven't studied functional analysis, and would prefer not to resort to a theorem I haven't read about to convince myself of this. If not, I guess I know what I need to learn next.

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Don't you mean "weakest topology"? – Mark Sep 27 '11 at 19:30
@Mark: The terms are synonymous. – Brian M. Scott Sep 27 '11 at 19:39
Isn't this just Banach-Alaoglu? – Jose27 Sep 27 '11 at 19:47

Developing Jose's comment:

We can identify $(\ell^1)^*$ with $\ell^\infty$ because every continuous functional $f$ on $\ell^1$ can be written as $f(x)=\sum_{k\geq 0}x_ku_k$ with some $(u_k) \in \ell^\infty$.

The coarsest topology on $\ell^\infty\equiv (\ell^1)^*$ such that all the mappings $(y_n) \to \sum_{k \geq 0} x_ky_k$ are continuous, where $(x_k)$ runs through $\ell^1$ is the weak* topology $\sigma((\ell^1)^*\equiv\ell^\infty,\ell_1)$.

The Banach-Alaoglu Theorem states that the unit ball is compact in the weak* topology and that is exactly what you ask.

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Thanks Beni. You don't happen to know a proof using only ideas of general topology do you? I haven't studied any functional analysis, so most of this is unfamiliar to me. – Gotye Sep 27 '11 at 22:47
@Gotye: A proof can be found in Brezis, Functional Analysis, Chapter 3. If you cannot find the book or the proof, I will post a proof. – Beni Bogosel Sep 27 '11 at 23:08
Thanks, I found the book, and am going to try to read through chapter 3 up to the result. – Gotye Sep 28 '11 at 1:33

As others have mentioned, this is a particular case of the Banach-Alaoglu theorem. But really that is a simple application of general topology, namely Tychonoff's theorem.
Let $P = \prod_{x \in \ell_1} [-\|x\|,\|x\|]$, which is compact by Tychonoff's theorem.
There is a natural embedding $\iota$ of your set $B = \{y \in \ell_\infty: \|y\| \le 1\}$ into $P$ by $\iota(y)_x = \sum_j y_j x_j$, and it is easy to see that this is a homeomorphism, because your topology exactly corresponds to the product topology. So all you need is to show that $\iota(B)$ is closed in $P$. But $\iota(B)$ consists of those members of $P$ that are linear, i.e. satisfy $p_{ax+by} = a p_x + b p_y$ for all $x, y \in \ell_1$ and $a,b \in \mathbb R$ (such a $p$ defines a member of $\ell_\infty$ by $y_j = p(e_j)$ where $e_j$ is the member of $\ell_1$ with $(e_j)_j = 1$ and all other elements 0). That is, $\iota(B)$ is the intersection of the closed sets $K(a,x,b,y) = \{p: a p_x + b p_y\}$ for $x, y \in \ell_1$ and $a,b \in \mathbb R$, so it is closed.

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Thanks Professor Israel, this seems to be more suited to my current level. – Gotye Sep 28 '11 at 1:34
Just to clarify, is $\|x\|=\sup_{n\in\mathbb{N}}|x_n|$ and similarly for $\|y\|$? I haven't seen this notation before. – Gotye Sep 28 '11 at 21:57
@Gotye: I think $\|\cdot\|$ means either $\ell_\infty$ norm or $\ell_1$ norm depending on what is inside: if $x\in\ell_1$, then $\|x\|=\|x\|_{\ell_1}=\sum_{n\in\mathbb{N}}|x_n|$ and if $y\in\ell_\infty$, then $\|y\|=\|y\|_{\ell_\infty}=\sup_{n\in\mathbb{N}}|y_n|$. – LostInMath Sep 30 '11 at 22:40

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