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Is it true that all compact subsets of $S^\infty:=\bigcup_{n=0}^\infty S^n$ are contained in some $S^n$?

I am interested in seeing how one can do this argument (if it is true) using "inductive arguments" (no functional analysis). The statement allows a fast proof that $S^\infty$ is weakly contractible: A map $S^m\to S^\infty$ has compact image and thus must be contained in some $S^n$, wlog make $n>m$ and then since $\pi_m(S^n)=0$ the image of the map must be contractible in $S^\infty$.

I think understanding this case would allow me to try to use similar arguments for other inductive limit topologies, like infinite joins etc.

s.harp
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  • If you define $S^n$ as the set of real sequences $(x_j){j\in \mathbb N}$ such that $x_j=0$ for $ j>n+1$ and $\sum{j=1}^{n+1}x_j^2=1$, and if the metric on $S^{\infty}$ is defined by $[; d((x_j)_j,(y_j)_j);]^2=\sum_j (x_j-y_j)^2$ then the Answer is NO. – DanielWainfleet Jun 27 '17 at 01:11
  • I think a counter-example would be ${(1,0,0,...)}\cup{x_n\mid n\in \Bbb N}$ where $x_n={(\sqrt{1-1/n}, 1/n,....,1/n,0,...,0)$. – s.harp Jun 27 '17 at 09:27
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    You usually give $S^{\infty}$ the final topology. That is, a subset $U$ is open if and only if $U \cap S^n$ is open in each $S^n$. With this topology, it is true that any compact subset is contained in $S^n$. It is in fact a particular case of a more general fact, that every compact subset of a CW-complex is contained in a finite subcomplex. – Goa'uld Jun 27 '17 at 17:17
  • @Goa'uld is my sequence and its limit point not a counter example to the statement though? – s.harp Jun 27 '17 at 20:35
  • That sequence is not even in $S^{\infty}$, their terms squared do not add up to one. – Goa'uld Jun 27 '17 at 21:37
  • @Goa'uld. We can replace $ \sqrt {1-1/n};$ in that example with $\sqrt {1-1/n^2};$ but as you said the final topology is not the same topology as the induced Hilbert-space metric topology – DanielWainfleet Jun 27 '17 at 21:59
  • For a more general statement, see https://math.stackexchange.com/questions/1584667/compact-subset-in-colimit-of-spaces – Eric Wofsey Jun 27 '17 at 23:19

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As I said in the comments, this could be done more in general, but I am going to show it for $S^{\infty}$ so you get the idea.

Suppose you have a compact subset $A$ of $S^{\infty}$. If $A$ is finite, then it is clearly contained in some $S^n$. So assume that $A$ is infinite and that it is not contained in any $S^n$.

Since $A$ is not contained in $S^1$, there is $x_1 \in A$ which is not in $S^1$. Say $x_1 \in S^{m_1}$. Since $A$ is not contained in $S^{m_1}$, there is $x_2 \in A$ which is not in $S^{m_1}$. In particular $x_1 \neq x_2$. Since $A$ is infinite we can continue this process and get a subset $B= \{ x_i \}$ of $A$ with $x_i \in S^{m_i} - S^{m_{i-1}}$ and where $m_1 < m_2 < m_3 < \ldots$ And in particular all the $x_i$ are different, so $B$ is infinite.

Note that $B \cap S^n$ is either empty or finite, hence $B \cap S^n$ is closed in $S^n$ for all $n$. Therefore $B$ is closed in $S^{\infty}$ and so it is closed in $A$. Since $A$ is compact, $B$ has to be compact.

Take any element $x_j$ of $B$. The complement $ C = B - \{x_j\}$ also satisfies that $C \cap S^n$ is closed in $S^n$ for all $n$. Therefore $C$ is closed in $S^{\infty}$, hence it is closed in $B$. So $\{ x_j \}$ is open in $B$. This means that $B$ has the discrete topology, every subset of $B$ is open.

Since $B$ has the discrete topology and it is compact, it has to be finite. But this is a contradiction with $B$ being infinite. Therefore $A$ had to be contained in some $S^n$.

Goa'uld
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