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

I am taking a topology course and we are now learning open and closed set. I am a bit confused to how to prove that a set is closed or opened, how should I approach these kind of problems. For example:

Question 1: Let $(\mathcal{X},d)$ be an arbitrary metric space. Prove that any set which contains a finite number of points $\{x_1,x_2,\ldots,x_n\}$ is closed.

Solution: If we take the point $x_i$ where $1\leq i\leq n$ and no matter how small we make $r$, in the ball some points are outside of our set. Hence the set is closed.

Question 2: Let $(\mathcal{X},d)$ be an arbitrary metric space. Prove that $B(\mathcal{X},r)=\{y\in\mathcal{X}:d(x,y)<r\}$ is open.

Solution: Take a point $y_0$ in the ball, and make a ball with radius $\frac{1}{2}(r-d(x,y_0))$, the all the points in this ball are in the actual ball in the question. Hence the ball in the question is open.

Are these proves right? I myself do not feel that I am proving anything. Could you please teach me the correct proof if these are not correct?


share|cite|improve this question
Q1 seems incorrectly formulated. It must not only contain finitely many points (every set does), but must be equal to a finite set. – wildildildlife Mar 12 '11 at 12:42
The solution to question 1 shows that the finite set is not open. This is not the same as closed. – Rasmus Mar 12 '11 at 15:09
As Rasmus said, one cannot stress enough that "not open" is not the same as "closed". A set can be both open and closed, it can be exactly one, or it can be neither. There are metric spaces where every set is both open and closed. – Jason DeVito Mar 12 '11 at 19:59
up vote 5 down vote accepted

For the first question: let's first prove that a one-point set is closed. Fix the point $x$, for the set $\{x\}$. Now consider any other point $y$. If $d(x,y)=\epsilon$, take $B(y,\epsilon/2)$. This open ball does not intersect $\{x\}$ (why?). This means that the complement of $\{x\}$ is open, which implies the set is closed. Now you know that any finite union of closed sets is closed (straight from the definition of topology by application of De Morgan's laws), and the result follows.

It isn't true that any ball around $x_i$, no matter how small, will contain points other than itself. To see this, consider the set of integers with $d(m,n)=|m-n|$.

For the second question, your approach is the right one, just don't forget (unless you didn't) to show that the ball you constructed is indeed contained in the original ball, by using properties of $d$.

So one way of determining whether a set is closed, is by looking at whether its complement is open, which in some cases could be easier.

share|cite|improve this answer

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.