# Subspace of infinite dimensional complex projective space generated by compact set

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.

• Just to be clear, what do you mean by the "projective subspace generated by $S$"? – Michael Albanese Apr 29 '16 at 13:58

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 $\mathbb{P}^{\infty}_{\mathbb{C}}$.
• 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 9:33