Let the distance of two subsets $\pmb S_1$ and $\pmb S_2$ of a given complete metric space $(X, d)$ be defined as $d(\pmb S_1,\pmb S_2)=\inf\{d(x,y):x\in\pmb S_1, y\in\pmb S_2\}$. Suppose that $\pmb S_1=\{x\}$ and $\pmb S_2$ is closed.
Prove $d(\pmb S_1,\pmb S_2)=d(x,y)$ for some $y\in\pmb S_2$.
In my opinion, if $d$ is not specified, how would I know whether the infimum is in the set or not because proving a set being closed involves $d$.
Does anyone have any idea?
If $X$ is a finite dimensional vector space, then the distance will be attained. For the proof consider $y \in \pmb S_2$ and $\pmb S_2^\prime = \pmb S_2 \cap \{z \in X \mid d(x, z) \le d(x,y)\}$. $\pmb S_2^\prime$ is compact and therefore $d(x ,\pmb S_2)$ is attained.
For an infinite dimensional vector space, the distance may not be attained. Consider $\pmb S_2 = \mathcal c_0$ the vector space of real sequences converging to zero equipped with the $\Vert \cdot \Vert_\infty$ norm. The hyperplane $H= \{(x_n) \in \mathcal c_0 \mid \displaystyle \sum_{n=1}^\infty 2^{-n} x_n =0\}$ is closed as it is the kernel of a continuous linear form. However for $a \notin H$, $d(a,H)$ is not attained.
For the proof, you can look here.