Contractibility of convex set

Question

Suppose that $\Omega$ is a convex open subset of an infinite dimensional vector space $E$ such that $\Omega$ is not contained in any finite dimensional subspace of $E$. Let $Q_m\subset \Omega$ denote a set of $m\in \mathbb{N}_{>0}$ distinct points.

Question: Is the space $\Omega\setminus Q_m$ contractible?

Added later:
Assume that for any set $Q_m'$ of $m$ distinct points $E\setminus Q'_m$ is contractible (I believe that this is true, see my answer below).
By "blowing up $\Omega$ like a balloon" we get a homeomorphism $\varphi$ from the convex open set $\Omega$ to $E$. More precisely, fix $x_0\in\Omega$ and for any open interval $I\subset \mathbb{R}$ let $\varphi_I:I\to \mathbb{R}$ be a (suitable) homeomorphism. Now, define $\varphi:\Omega \to E$ by $$\varphi(x):=\begin{cases}\varphi_{\Omega\cap \mathbb{R}\cdot (x-x_0)}(x), \text{ if } x\neq x_0 \\ x_0 , \text{ if } x= x_0.\end{cases}$$ (I know that the definition of $\varphi$ is not formally correct but I think that the idea is clear.) Restricting $\varphi$ to $\Omega\setminus Q_m$ we get a homeomorphism between $\Omega\setminus Q_m$ and $E\setminus \varphi(Q_m)$. By assumption $E\setminus \varphi(Q_m)$ is contractible and hence $\Omega\setminus Q_m$ is also contractible.

Is the intuition behind this heuristic argument any good? Can it be made into a complete and rigorous argument?

Answer 1

Following the suggestion of Martin I'll post an answer to my own question. I would appreciate some feedback. Please let me know if you notice any errors.

Let $Q'_m\subset E$ be a set of $m\in \mathbb{N}$ distinct points.

Claim: $E\setminus Q'_m$ is contractible.

Proof: Write $E$ as $E\cong W\oplus V$ where $W\subset E$ is a finite dimensional subspace containing $Q'_m$. Let $\Phi:W\oplus V \to W\oplus V$ be a linear map such that $\Phi|_W=id_W$ and $\Phi|_V$ has no non-zero eigenvalues. (For example $\Phi|_V$ could be chosen as $(v_1,v_2,v_3,\ldots)\mapsto (0,v_1,v_2,v_3,\ldots)$.) The map $\Phi$ gives rise to an isomorphism from $E$ to a genuine subspace $im(\Phi)\subsetneq E$. It also induces a map $\Phi:E\setminus Q'_m\to E\setminus Q'_m$.
Now, $\Phi_t:E\setminus Q'_m\to E\setminus Q'_m$ defined by $$\Phi_t(x):=(1-t)x+t\Phi(x)\quad t\in [0,1]$$ is a homotopy from $id_{E\setminus Q'_m}$ to $\Phi$.
Pick $x_0\notin im(\Phi)\cup Q'_m$. Then $\Psi_t:E\setminus Q'_m\to E\setminus Q'_m$ given by $$ \Psi_t(x):=(1-t)\Phi(x)+tx_0\quad t\in [0,1]$$ is a homotopy from $\Psi_0=\Phi$ to the constant map $\Psi_1\equiv x_0$. Hence, $E\setminus Q'_m$ is contractible.$\qquad\square$

This claim together with the fact that $\Omega\setminus Q_m$ is homeomorphic to $E\setminus Q'_m$ (see the comments by Martin and my argument in the question) shows that $\Omega\setminus Q_m$ is contractible.

I'd like to thank Martin for his helpful comments.