I'm generalizing some theorems to Polish metric spaces, which were originally proved for $\mathbb R^n$. Hence, as I'm adapting the proofs, I'm checking whether where I can reuse the same arguments. In one of my proofs, I had the following:
"Define $\pi_R : \mathbb R^n \to B(R)$ to be the projection operator on the closed ball with radius $R$ and centered at zero".
Now I'd like to use this for an $X$ metric Polish space. Can I "safely" assume that there exists such projection operator $\pi_R: X \to B(R)$ such that the closed ball is centered at some $x_0 \in X$? How can I know that this map is well defined?
The answer is "no". Consider the separable real Banach space $\ell^2$. Any closed subspace of $\ell^2$ is Polish, a fact we shall soon use.
Let $e_n$ be the $n$th basis vector, i.e. $e_n$ is the sequence such that the $n$th entry in the sequence is $1$, while the other entries are $0$. Let $$C = \left\{\frac{1 + \frac{1}{n}}{4}e_{n+1} : n \ge 1\right\}$$ and $$X = C \cup \{2e_1\}.$$ Note that every point in $X$ is isolated, and hence $X \subseteq \ell^2$ is closed, and hence a Polish space in its own right. If we have distinct $\frac{n + 1}{2n} e_n$ and $\frac{m + 1}{2m} e_m$ in $C$, then $$\left\|\frac{n + 1}{4n} e_n - \frac{m + 1}{4m} e_m\right\|^2 = \left(\frac{n + 1}{4n}\right)^2 + \left(\frac{m + 1}{4m}\right)^2 \le \frac{1}{2} + \frac{1}{2} = 1,$$ but $$\left\|\frac{n + 1}{4n} e_n - 2e_1\right\|^2 = \left(\frac{n + 1}{4n}\right)^2 + 4 \ge 4.$$ Therefore, $C$ is the closed unit ball in $X$, centred at any point in $C$. We can now show that $2e_1$ has no closest point, i.e. the projection map simply does not extend to this point in $X$. We see that
$$\left\|\frac{n + 1}{4n} e_n - 2e_1\right\|^2 = \left(\frac{n + 1}{4n}\right)^2 + 4 \to \frac{17}{4},$$
while remaining strictly greater than $\frac{17}{4}$ for every $n \ge 2$. Therefore, there is no point in $C$ that achieves the infimal distance of $\frac{\sqrt{17}}{2}$ to any unit ball centred in $C$.