# some confusion about singleton set?

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

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.

• I think...the OP tried to show that $\{0\}$ is not open, but this should have been balls with centre $0$ instead of $x>0$...(and of course not open does not imply closed) – Calvin Khor Feb 21 at 15:31
• @CalvinKhor You are probably right. – José Carlos Santos Feb 21 at 15:35

"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.

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.