some confusion about singleton set?

Question

Is the set $\{0\}$ is closed in $(\mathbb{R} , |.|)$ ? where $|.|$ denotes the usual metric on $\mathbb{R}$

My attempt : yes , because i think $\{0\}$ is not open for all $x> 0, (x- \epsilon, x + \epsilon) \notin \{0\}$

Is its correct?

Any hints/solution

Answer 1

No, it is not correct. Asserting that $\{0\}$ is closed in $\mathbb R$ means that $\mathbb{R}\setminus\{0\}$ is open in $\mathbb R$. That is what you should try to prove (and it is not hard). But the assertion “$\{0\}$ is not open for all $x>0$, $(x-\varepsilon,x+\varepsilon)\notin\{0\}$” makes no sense.

Answer 2

"Not open" does not imply closed. What you need to show is that the complement $\mathbb{R} \setminus \{0\}$ is open, e.g. by writing $\mathbb{R} \setminus \{0\}$ as a union of open intervals.

Answer 3

Take any sequence in $\{0\}$. Then it must be the zero sequence and thus converges to 0, which lies in the set. So $\{0\}$ is closed.