Show that homeomorphism is an equivalence relation in metric spaces It needs to be shown that homeomorphism is reflexive, symmetric and transitive in all metric spaces.
Reflexivity seems to be easy to show, but I'm not sure how to do the rest. Any help?
 A: The relation here is “$X$ is homeomorphic to $Y$”.
To show this relation is symmetric, you need to show that if $X$ is homeomorphic to $Y$, then $Y$ is homeomorphic to $X$.  So if you have a homeomorphism $f:X\to Y$, you want to show that there must be a homeomorphism $h: Y\to X$.  Can you think of any way of constructing the desired homeomorphism $h$ from the the one you already have, namely $f$?
To show this relation is transitive, you need to show that if $X$ is homeomorphic to $Y$, and $Y$ is homeomorphic to $Z$, then $X$ is homeomorphic to $Z$.  So if you have a homeomorphism $f:X\to Y$, and a homeomorphism $g:Y\to Z$, you want to show that there must be a homeomorphism $h: X\to Z$.  Can you think of any way of constructing the desired homeomorphism $h$ from the the ones you already have, namely $f$ and $g$?
In each of the two parts, there are two steps.  First, try to guess what $h$ might  be. (In each part there's really only one possible choice.)  Then show that the $h$ that you guessed is actually a homeomorphism. 
