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Let's define infinite projective space space $\mathbb{CP}^{\infty}$ as direct limit $\lim \limits_{\rightarrow} \mathbb{CP}^n$. In a class I attended it was claimed that every continuous map $S^k \to \mathbb{CP}^{\infty}$ is actually valued in some subspace $\mathbb{CP}^N$ for some sufficiently high $N$. How to prove this?

Blazej
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  • A subspace of a CW-complex is compact iff it is closed and it intersects finitely many cells, see for example this question – Alessandro Codenotti Jan 31 '19 at 19:01
  • If $f_i: X_i \to X_{i+1}$ is a sequence of inclusions of compact Hausdorff spaces, then the direct limit topology on the union is such that any compact subset lies in one of the $X_i$. This is a nice exercise with the definitions. –  Jan 31 '19 at 19:06
  • @MikeMiller this is precisely what I expected, but I had some trouble with verifying this property. Could you prove some additional hints? – Blazej Jan 31 '19 at 20:04

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Here's a very general statement which is enough for your situation:

Let $X_i \to X_{i+1}$ be a sequence of embeddings of $T_1$ spaces. Then any compact subset $K\subset \mathrm{colim}_i X_i$ is actually contained in an $X_i$.

A proof can be found here : Compact subset in colimit of spaces

It's a very general statement, and the proof is not hard to follow.

Maxime Ramzi
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  • As pointed out in a comment, my argument was false. I have deleted it. –  Feb 01 '19 at 18:26
  • @MikeMiller : thanks for pointing it out, I had not checked the argument, only the statement – Maxime Ramzi Feb 01 '19 at 18:30
  • Thank you! I hoped for something exactly of this type, but I expected much stronger assumptions to be necessary, e.g. each $X_i$ compact Hausdorff. – Blazej Feb 01 '19 at 21:42