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Let $$ X= \{u=(u_1, u_2, \ldots): u_n \ne 0 \text{ only for a finite number of terms}\}\subseteq\mathbb R^\mathbb N, $$ with the topology inherited from $\mathbb R^\mathbb N$ (the "pointwise convergence topology"). Find the bounded sets and prove they're compact.

I show you what I've done. First of all, I think - but I'm not sure on how to prove it - the topology of $X$ is equivalent to the one generated by the (countable) family of seminorms $$ p_n(u)=\vert u_n\vert, \qquad n \in \mathbb N. $$ This allows me also to write an "explicit" form of an invariant metric $d$, e.g. $$ d(x,y) = \sum_{n \in \mathbb N} 2^{-n}\frac{p_n(x-y)}{1+p_n(x-y)} $$ Do you agree? Have you got any idea about how I can prove that the two topologies are the same? Is it true?

Anyway, let us come back to bounded sets. If my idea is correct, bounded sets of $X$ are exactly the ones for which every $p_n$ is bounded (as real valued functions). Is there a more explicit description?

And what about compactness? I suppose I have to use some theorem about compactness (Tychonov?) but I can't see how.

Thanks in advance.

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What does bounded mean here ? bounded to which metric ? –  Amr Dec 27 '12 at 17:11
    
@Amr: "boundedness" is not a metric concept. It can be defined for arbitrary topological vector spaces. A set $E \subset X$ ($X$ topological vector space) is bounded iff for every neighbourhood $W$ of $0_X$ there exists a $t>0$ s.t. $E \subseteq tX$ (or, equivalently $E \subset sX$ for every $s>t$). Hope now it is clear. Thanks for pointing it out, anyway. –  Romeo Dec 27 '12 at 17:18
    
I think that you meant $E\subset tW$ –  Amr Dec 27 '12 at 17:41
    
Of course, sorry for the typo. –  Romeo Dec 27 '12 at 22:16
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1 Answer

up vote 1 down vote accepted

For $r>0$ and $n\in\Bbb N$ let $B_n=\{x\in X:|x_k|<2^{-n}\text{ for all }k\le n\}$; $\{B_n:n\in\Bbb N\}$ is a local base at $z=\langle 0,0,0,\dots\rangle\in X$. Thus, a set $K\subseteq X$ is bounded iff for each $n\in\Bbb N$ there is a $t_n>0$ such that $K\subseteq t_nB_n$. Clearly this is the case iff $\pi_n[K]$ is bounded in $\Bbb R$ for each $n\in\Bbb N$, where $$\pi_n:X\to\Bbb R:x\mapsto x_n$$ is the canonical projection map. Since your $p_n(x)=|\pi_n(x)|$, this is clearly equivalent to the characterization that you found.

It is not true that every bounded subset of $X$ is compact: the bounded set $$I=\{x\in X:x_0\in(0,1)\text{ and }x_k=0\text{ for all }k>0\}$$ is clearly homeomorphic to $(0,1)$ with the usual topology and hence not compact. It isn’t even true that every closed, bounded set in $X$ is compact.

To see this, for $n\in\Bbb N$ let $x^n\in X$ be defined by

$$x^n:\Bbb N\to\Bbb R:k\mapsto\begin{cases} 1,&\text{if }k\le n\\ 0,&\text{if }k>n\;, \end{cases}$$

and let $K=\{x^n:n\in\Bbb N\}$; clearly $K$ is bounded. If $x\in X\setminus K$, then either $x_n\notin\{0,1\}$ for some $n\in\Bbb N$, or there are $m,n\in\Bbb N$ such that $m<n$, $x_m=0$, and $x_n=1$. In the first case let $U$ be an open nbhd of $x_n$ in $\Bbb R$ such that $0,1\notin U$; then $\pi_n^{-1}[U]$ is an open nbhd of $x$ in $X$ that is disjoint from $K$. In the second case let $U$ and $V$ be disjoint open nbhds of $0$ and $1$, respectively, in $\Bbb R$; then $\pi_m^{-1}[U]\cap\pi_n^{-1}[V]$ is an open nbhd of $x$ in $X$ that is disjoint from $K$. Thus, $K$ is closed in $X$. (In fact the only cluster point of $K$ in $\Bbb R^{\Bbb N}$ is $\langle 1,1,1,\dots\rangle$, which of course is not in $X$.)

Now let $U$ and $V$ be disjoint open nbhds of $0$ and $1$, respectively, in $\Bbb R$. For $n\in\Bbb N$ let $$W_n=\pi_n^{-1}[V]\cap\pi_{n+1}^{-1}[U]\;,$$ and let $\mathscr{W}=\{W_n:n\in\Bbb N\}$; clearly $\mathscr{W}$ is an open cover of $K$ with no finite subcover.

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Thanks a lot for your beautiful and detailed answer. I'm very surprised by the easy counterexample to compactness of bounded sets you've found; the text of the exercise is wrong. Thanks again for your help. –  Romeo Dec 27 '12 at 22:19
    
@Romeo: You’re very welcome. –  Brian M. Scott Dec 28 '12 at 7:31
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