Distance attained by a function

Question

Let $A$ be a subset of $\mathbb R^n$ and let $x\in \mathbb R^n$. Then $\exists y_0\in A$ such that $d(x,y_0)=d(x,A)$ if

1. $A$ is a non-empty subset of $\mathbb R^n$.

2. $A$ is a non-empty closed subset of $\mathbb R^n$.

3. $A$ is a non-empty compact subset of $\mathbb R^n$.

4. $A$ is a non-empty bounded subset of $\mathbb R^n$.

In my opinion, option 3. is correct. If we define a function $f:A\to \mathbb R$ by $f(y)=d(x,y)$ for all $y\in A$, then $f$ is continuous (in fact uniformly continuous) and so attained its bounds if $A$ is compact. Since $d(x,A)=\inf\{d(x,y): y\in A\}$, there fore there is $y_0\in A$ such that $f(y_0)=d(x,A)$. But is the option 2 is wrong? I could not find any clue. Please help to find the answer.

Answer 1

Hint: Construct a sequence of $y_{n}\in A$ with the following property, for each $y_{n}$, $d(y_{n}, x)\leq d(A, x)+\frac{1}{n}$. Now, does $y_{n}$ converge? can you adjust it if necessary to make it converge? does it converge to something in $A$?

Hints on specific numbers:

1) What if $A=\mathbb{Q}$ and $x\notin \mathbb{Q}$?

2) If $A$ is closed then $A\cap \{y: d(x, y)\leq d(x, A)+\frac{1}{n}\}$ is a compact set.

3) Compact implies sequentially compact...

4) What if $A=\mathbb{Q}\cap [0, 1]$ $x\in (0, 1)\setminus \mathbb{Q}$.