Let $A$ be a closed subset of the interval $[0,1]$. Show that the quotient space $$W=(-2,2)\big/A$$ is Hausdorff.
I guess we need to explicitly find the disjoint open sets for each two $x\neq y$ from $W$ depending on where their inverse images are (in $A$ or in $[0,1]-A$). Could you please help me formulate the proof?
On
I am working on topology as well, so my answer is likely flawed, although maybe you can draw some ideas from it..
To show it is Hausdorff, we need that for arbitrary distinct points $x+A\in W$ and $y+A\in W$, $\exists U,V$ neighborhoods of $x+A$ and $y+A$ such that $U\cap V=\emptyset$
Consider the quotient map to be $q:(-2,2)\to (-2,2)/A$ defined by $q(x)=x+A$
Assume to the contrary that for arbitrary $x+A$ and $y+A$, $U\cap V\neq\emptyset$ for all neighborhoods. In particular, we can construct a net $(z_{\lambda}+A)_{\lambda\in\Lambda}$ in $W$ that converges to both $x+A$ and $y+A$
Note that the preimage under $q$ of the elements of this net is a net of representatives $(z_{\lambda})_{\lambda\in\Lambda}$ in $(-2,2)$, which converges simultaneously some representative $q^{-1}(x+A)$ of $q^{-1}(U)$ and some representative $q^{-1}(y+A)$ of $q^{-1}(V)$. But since $x+A\neq y+A$, we have that $q^{-1}(x+A)\neq q^{-1}(y+A)$ (or else they would have been in the same equivalence class). So that the net $(z_{\lambda})_{\lambda\in\Lambda}$ converges to two different points simultaneously in $(-2,2)$
This is a contradiction to the fact that $(-2,2)$, equipped with the usual subspace topology on $\mathbb{R}$, is Hausdorff
If $X$ is a regular space (i.e. $T_3$) and $A$ a closed subset of $X$, then identifying $A$ to a point, which means taking the quotient wrt the equivalence relation $R_A$ that has $A$ and $\{x\}, x \notin A$ as its classes, gives a Hausdorff space as its quotient, oft denoted $X/A$, with quotient map $q: X \to X/A$.
This holds, as two classes are either of the form $\{x\}$ and $\{y\}$ (both not in $A$) and so we separate them with open disjoint neighbourhoods $U_x$ and $U_y$ both disjoint from $A$ (which can be done as $A$ is closed) and then $U_x$ and $U_y$ are saturated under $R_A$ so that $q[U_x]$ and $q[U_y]$ are open in $X/A$. Otherwise we want to separate the classes $\{x\}$ ($x \notin A$) and $A$ which we can do in $X$ by $T_3$-ness and again the images under $q$ are as required.
The OP's problem is a special case as $[0,1]$ is closed (even compact, in which case we only need $X$ to be $T_2$ in general) and $(-2,2)$ is metric so certainly has all the separation axioms anyone needs..