0

This question is similar to this one, but with the infinite dimensional complex space instead of the complex separable Hilbert space.

My question is: if $S\subseteq \mathbb C P^\infty $ is a compact subset, then is it true that the projective subspace generated by $S$ is finite dimensional?

The counterexample in the linked question clearly fails in this modified case.

Community
  • 1
user335658
  • 1

1 Answers1

0

The answer is no!

Example: Similarly to Hilbert cube in $\mathbb{C}^{\infty}$, let $H=\{0\}\times[0,1]\times\left[0,\frac{1}{2}\right]\times\dots$ it is a compact subset of $\mathbb{C}^{\infty}$, therefore $\pi(H\setminus\{\underline{0}\})=K$ is a compact subset of $\mathbb{P}^{\infty}_{\mathbb{C}}$, where $\pi:\mathbb{C}^{\infty}\setminus\{\underline{0}\}\to\mathbb{P}^{\infty}_{\mathbb{C}}$ is the canonical projection; but the projective (linear) space generated by $K$ is the hyperlane $\{x_0=0\}$, where $x_0$ is the first coordinate in $\mathbb{P}^{\infty}_{\mathbb{C}}$.

Armando j18eos
  • 3,907
  • Your subset $H$ contains zero, which is not in the domain of $\pi$. Perhaps you meant to make the first coordinate equal to $1$? – Dylan Thurston Mar 12 '18 at 22:49
  • Yes it is, $\underline{0}\in H$; I remark that $H\setminus{\underline{0}}=K$ is a compact subset of $\mathbb{C}^{\infty}\setminus{0}$, so $\pi(K)$ is a compact subset of $\mathbb{P}^{\infty}_{\mathbb{C}}$ and it generates ${x_0=0}$. Are you agree? – Armando j18eos Mar 14 '18 at 09:33