# Identification of a Subset to a Point

If $$A$$ is a subspace of a topological space $$S$$, we can define a relation $$∼$$ on S by declaring $$x ∼ x\quad\text{for all}\quad x\in S$$ (so the relation is reflexive) and $$x ∼ y\quad\text{for all}\quad x, y\in A.$$

Question 1. On the book it says that this is an equivalence relation on $$S$$. Why?

Question 2. Who are the equivalence classes?

Thanks!

• Short 1) $x,y,z \in A$ – Mann Apr 15 at 13:31
• I am not sure about the downvote, but I guess it is because you didn't show "efforts" in your question, for example by explaining which axiom of equivalences you have problems to check in Question 1. – Taladris Apr 15 at 13:39
• Just think for a bit, and handle cases. If $x \sim y$ then either $x=y$ and symmetry is obvious, or $x \neq y$ and then $x, y \in A$. But if $x$ and $y$ both live in $A$, then $y$ and $x$ both live in $A$.... – Randall Apr 15 at 13:48
• You should mention it in your question I think. Symmetry is obvious: if $x\sim y$, then either $x=y$ or ($x$ and $y$ are in $A$). This implies that $y=x$ or ($y$ and $x$ are in $A$), so $y\sim x$. For transitivity, it is no more difficult, but you need to consider cases. – Taladris Apr 15 at 13:48
• @JackJ. You would say that if $x \in A$ then $[x]=A$ and if $x \in X - A$ then $[x]=\{x\}$. – Randall Apr 15 at 19:12

• If $$x \sim y$$ then either we're in the case $$x \sim x$$ (option 1) and so $$y=x$$ and $$y \sim x$$ is again given. Or we're in the case $$x,y \in A$$ and so then also $$y \sim x$$. The statement $$x\in A \land y \in A$$ is symmetric in $$x$$ and $$y$$.
• For proofs of transitivity we can always assume WLOG that all three points involved are different: in this case $$x \sim y$$ and $$y\sim z$$ then implies $$x,y\in A$$ and $$y,z\in A$$, so clearly also $$x,z\in A$$ (we jus state less info) so $$x \sim z$$.
The equivalence classes of course are $$A$$ and all $$\{x\}$$ where $$x \notin A$$. So $$A$$ becomes a new point in the quotient space and all other are left untouched.