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Let $\, X := \mathbb{R}^3\Big/_{\sim} \,$ where $\, \sim \,$ is defined as: $\,x \sim y \iff x = y \quad \lor \quad \lVert x\rVert = \lVert y \rVert > 4$. Say wheter the canonical map $\, \pi :\mathbb{R}^3 \rightarrow X$ is open or not and if $X$ is a second-countable space or not.

If $\pi$ is open then we can easily say that $X$ is a second-countable space and to me it seems, intuitively, that $\pi$ should be open, but i can't make any progress towards building a formal proof of this (neither find a counter-example). Any hints would be appreciated.

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Hint

$\pi$ is open if and only if for any open subset $U \subset \mathbb R^3$ $\pi^{-1}(\pi(U))$ is open.

Now you can prove (by considering the cases $\Vert x \Vert <, =, > 4$) that for an open ball $B(x,r)$ centered on $x$ of radius $r < \Vert x \Vert$, $\pi^{-1}(\pi(B(x,r)))$ is open. Which proves that $\pi$ is open.

The "interesting case" is when $B(x,r)$ intersects the sphere centered on the origin of radius $4$. In particular if $\Vert x \Vert = 4$, you can prove that $$\pi^{-1}(\pi(B(x,r))) = B(x,r) \cup \left(B(0,4+r) \setminus \overline{B}(0,4)\right)$$ which is the union of two open subsets.

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Yes, $\pi$ is open.

Let $K = \{x \in \mathbb{R}^3 : \|x\| > 4\}$. For $S \subseteq \mathbb{R}^3$ we have

$$\pi(S) = \{[x] : x \in S\setminus K\} \cup \{S(0, \|x\|) : x \in S \cap K\}$$ where $S(0,r)$ is the sphere of radius $r$ centered at origin.

Therefore

$$\pi^{-1}(\pi(S)) = (S \setminus K) \cup \bigcup_{x \in S \cap K} S(0, \|x\|) = S \cup \bigcup_{x \in S \cap K} S(0, \|x\|)$$

because $S\cap K \subseteq \bigcup_{x \in S \cap K} S(0, \|x\|)$.

If $S$ is open, then $S \cap K$ is also open so $S \cap K = \bigcup_{i \in I} B(x_i, r_i)$ for some family of open balls $(B(x_i, r_i))_{i \in I}$. Therefore $$\bigcup_{x \in S \cap K} S(0, \|x\|) = \bigcup_{i \in I} \{x \in \mathbb{R}^3 : \|x_i\| - r < \|x\| < \|x_i\| + r\}$$ is open as a union of open sets. Hence $\pi^{-1}(\pi(S))$ is open.

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