Is the subset of a relatively compact set relatively compact? I am trying to read a paper and there is something I do not understand. 
Given a compact set $K \in \mathbb{R}^m$ and a set $A \subset K $ and a compact operator $T:\mathbb{R}^m \to \mathbb{R}^m $, I am trying to determine if $A-T(K) $ which I think is defined as 
$$A-T(k)= \bigcup_{x \in K} A-T(x) $$ where $A-T(x)$ is translating all point in $A$ by $T(x)$. This is not explicitly mentioned in the paper so its what I assume.
Now clearly $A$ being a subset of a compact set , its relatively compact and since $T$ is a compact operator , $T(K)$ is relatively compact .
Can I also say that $A-T(K)$ defined above is relatively compact?
I tried to argue using the sequential  compactness definition but couldn't prove it .
If the above is true can I conclude as in the compact case that its subset is relatively compact? Why?
 A: Yes, $A-T(K)$ is relatively compact in $\mathbb{R}^{m}$. $T$ is a bounded operator, so the image of any bounded subset of $\mathbb{R}^{m}$ under $T$ will be relatively compact. Therefore, it follows from $K \subset \mathbb{R}^{m}$ being compact that $T(K)$ is relatively compact. Any subset of a compact set is at least relatively compact, which means that $A\subset K$ is guaranteed to be relatively compact. All you need to show $A-T(K)$'s relative compactness is that cl$(A-T(K))$ is bounded, which clearly follows from the fact that both $A$ and $T(K)$ are both relatively compact.
Regarding $A-T(x)$ for a particular $x \in K$, consider what it would mean if $A-T(x)$ was not relatively compact. It would imply that $T(x)$ was unbounded for some $x \in K$--a compact space. This directly contradicts the fact that $T$ is a bounded operator, so it can't be the case. We now have that for all $x \in K, \ T(x)$ is bounded. Combine this with the fact that $A$ is bounded, and it follows that for all $x \in K$, cl$(A - T(x))$ is bounded, so $A-T(x)$ is relatively compact. 
