Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

let $(X,d)$ be a metric space. How I can show that any finite subset of $X$ is closed.

Can a finite subset of $X$ be open ?


  • a set $F\subseteq X$ is closed (in$(X,d)$) if $\bar F =F$.
  • a set $U\subseteq X$ is open (in$(X,d)$) if $U^o=U$
share|cite|improve this question
Hint: Can you show that any set containing just one element is closed? For the second question, consider the discrete metric. – Tobias Kildetoft Mar 20 '13 at 17:40
For such basic questions it will help answerers if you include the precise definitions you have been given. What is your definition of an open set? A closed set? – Pete L. Clark Mar 20 '13 at 17:42
For the second question consider a finite metric space (just take $\{0,1,2\}$ with $\lvert \cdot-\cdot\rvert$ for example) – Stefan Mar 20 '13 at 17:54
Another way to read your definition of closed is that a set is closed in $(\mathbb{X},d)$ if and only if it contains all of its limit points. How many limit points does a finite set have? – Todd Wilcox Mar 20 '13 at 18:01
up vote 7 down vote accepted

For an approach even more basic than Andrew Salmon’s, let $\langle X,d\rangle$ be a metric space, and let $F$ be any finite subset of $X$. The empty set is closed by definition, so we might as well assume that $F\ne\varnothing$. Now suppose that $x\in X\setminus F$, and let $r_x=\min\{d(x,y):y\in F\}$. Then $r_x>0$ (why?); what can you say about $B(x,r_x)$, the open ball of radius $r_x$ centred at $x$?

Yes, a finite set in a metric space can be open. First, the empty set is always open. Other than that, though, it depends on the space. No finite, non-empty subset of $\Bbb R^n$ is open, for instance, for any $n\in\Bbb Z^+$. However, if $X$ is any set at all, the function $d:X\times X\to\Bbb R$ defined by

$$d(x,y)=\begin{cases}1,&\text{if }x\ne y\\0,&\text{if }x=y\end{cases}$$

is a metric, often called the discrete metric, and every subset of $X$ is open.

share|cite|improve this answer
answering (why?) :$r_x>0$ because $F\cap X/F=\varnothing$.(Is this make any sense?) – Jhwana Mar 20 '13 at 18:22
$B(x,r_x)$ is contained in $X/F$ thus $X/F$ is open , so $F$ is closed. – Jhwana Mar 20 '13 at 18:27
@Fayz: I’m afraid not. Perhaps it will be easier if I let $F=\{x_1,\dots,x_n\}$ for some finite $n$. Then $r_x$ is the minimum of the numbers $d(x,x_k)$ for $k=1,\dots,n$. Since $x\notin F$, each $d(x,x_k)>0$. And there are only finitely many of these positive numbers, so there is a smallest one, which I call $r_x$. But your second comment is right on the money. – Brian M. Scott Mar 20 '13 at 18:27
Thank you so much – Jhwana Mar 20 '13 at 19:44
@Fayz: You’re very welcome. – Brian M. Scott Mar 20 '13 at 19:46

For an answer that doesn't require any knowledge of separation axioms, consider the limit points of $\{ x \}$. Fix a point $y$ not in this set. Can $y$ be a limit point (hint: remember that $d(x,y) \ne 0$)?

Now remember that the finite union of closed sets is closed.

share|cite|improve this answer

Metric spaces are hausdorff spaces (T2) and therefore T1. This means that points are closed and a finite union of closed sets, as is well known, is closed.

share|cite|improve this answer
Good answer, +1 – Rustyn Mar 20 '13 at 17:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.