I am a little bit confused. Assume that $Y$ is a subset of $X$, with the inherited topology from $X$. What does it mean the $K\subset Y$ is a relatively compact subset of $Y$?
One option, is that $K= C\cap Y$, where $C$ is a relatively compact subset in $X$.
Another option, is the $\overline{K}\cap Y$ is compact in $Y$.
To be specific, I am looking on the following example:
Let $X$ be a topological (locally compact Hausdorff), $\{D_n\}_{\ n\in\mathbb{Z}}$ be a sequence of open subsets inside $X$. For every $n$, let $T_n: D_n\to D_{-n}$ be a homeomorphism. Consider the set $Y:=\{(T_n(x),n,x)\in X\times \mathbb{Z}\times X| x\in D_n\}$ with the inherited topology from $X\times \mathbb{Z}\times X$.
I want to say that if $K\subset Y$ is relatively compact, then $K$ is contained in a set of the form $\{(T_n(x),n,x)\in X\times \mathbb{Z}\times X| x\in D_n\cap C, \ n\in [-N,N]\}$ for some relatively compact set $C$ in $X$ and $N\in\mathbb{N}$. Is it true?
Thanks!
According to definition a subset of a topological space is relatively compact if its closure (in that space) is compact.
So I suspect that $K\subseteq Y$ is a relatively compact subset of $Y$ it is closure in $Y$ is compact.
Further if $\overline K$ denotes the closure of $K$ in original space $X$ then the closure of $K$ as a subset of $Y$ equals $\overline K\cap Y$.
This results in the second option that you mention:$$K\text{ is a relatively compact subset of }Y\subseteq X\text{ iff }\overline K\cap Y\text{ is compact}$$
I advice you not to accept this answer, but just think it over. Maybe a real topologist will pass by to adjust.