I came across the following assertion: Let $X$ be a metric space. Let $Y\subset X$ be a bounded set. Then, there exists a smallest compact set $K$ in $X$ such that $K\supset Y$.
How can I prove this? In $\mathbb{R}^n$, this looks trivial, because a compact set is equivalent to a closed and bounded set, so $K$ is just a closure of $Y$. But, in general metric space, I'm not sure how to prove this result.
Let $X$ be a metric space. Let $Y\subset X$ be a bounded set. Then, there exists a smallest compact set $K$ in $X$ such that $K\supset Y$.
This assertion is not true in general.
For an example, let $(V,||.||)$ be a normed vector space with dim$(V)$ $\geq \infty$. Note that the norm on the vector space induces a metric $d$ on the vector space as follows:
\begin{equation*} d(u,v) = ||u-v|| \hspace{1cm} \forall u,v \in V \end{equation*}
Let $B[0,1]$ denotes the close unit ball of $V$. i.e:
\begin{equation*} B[0,1] = \{u \in V: ||u|| \leq 1\} \end{equation*}
By definition it is a bounded set. If the assertion was true then there exists a smallest compact set $K$ in $V$ such that $B[0,1] \subset K$. But close subsets of compact set in a metric space is compact. So this forces $B[0,1]$ to be compact. But it is a well known fact that:
The closed unit ball in a normed linear space is compact if and only if the vector space is finite dimensional.
You will get a discussion on this statement in this link. But we have taken $V$ to be infinite dimensional. So $B[0,1]$ is not compact. This is a contradiction.
