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?


  • $\begingroup$ Short 1) $x,y,z \in A$ $\endgroup$ – Mann Apr 15 at 13:31
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    $\begingroup$ 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. $\endgroup$ – Taladris Apr 15 at 13:39
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    $\begingroup$ 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 $A$.... $\endgroup$ – Randall Apr 15 at 13:48
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    $\begingroup$ 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. $\endgroup$ – Taladris Apr 15 at 13:48
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    $\begingroup$ @JackJ. You would say that if $x \in A$ then $[x]=A$ and if $x \in X - A$ then $[x]=\{x\}$. $\endgroup$ – Randall Apr 15 at 19:12

Check the definitions to see it is an equivalence relation:

  • reflexivity is given.
  • 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.


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