# The existence of a limit point of a closed set

Walter Rudin define a closed set as:

2.18 (d) A subset E of a metric space is closed if every limit point of E is a point of E.

I don't see in this definition nothing about the existence of a limit point of such set E. It's like the definition of a compact set, I don't see nothing about the existence of a open cover of such set, but I know that there is at least one open cover, its own metric space.

For me, the definition of a closed set doesn't guarantee the existence of a limit point, ¿am I right or not? and ¿Why?

• You are right. If the set doesn't have limit points then it is automatically closed. – dafinguzman Oct 11 '15 at 16:36

The empty set is a closed set with no limit points. Other than that any discrete set also has no limit points. But, these are closed for the same reason the empty set is closed. Because they do contain all of their limit points, which is none.

• Speaking of things that are trivially closed can sometimes yield weird statements, like "they do contain all of their limit points, which is none." I think it would be a little more clear in this case to say something like "These sets are closed because there are no limit points of the set contained outside the set" – graydad Oct 11 '15 at 16:42
• That's definitely a clearer statement. But I guess I was trying to emphasize that it's not necessary to have any limit points. – Konrad Wrobel Oct 11 '15 at 16:44
• For sure! I agree with your answer, and your statement was absolutely correct. But all the same I think we should always try to make math as clear as possible when we are able :) – graydad Oct 11 '15 at 16:49
• I think you say that the empty element is a limit point of every set, but if it is true, then I can say that the empty element is belong in the complement of such set. So, the empty element is outside of the set... I think: If a set has not limit point, then it's closed by unanimous agreement? or I just don't understand you? – Yesid Fonseca V. Oct 11 '15 at 17:17
• What I mean to say is that if you have no limit points, then you have nothing to check and your set is closed. Examples of such sets are the empty set or discrete sets like, {1} or {1,2} or {1.5, 10, 300}. The way the first comment states it is somewhat more clear. The definition of containing all limit points is equivalent to having no limit points outside the set. – Konrad Wrobel Oct 11 '15 at 17:25

No, the definition does not imply any limit points exist. If a set does not have any limit points at all, then all zero of its limit points are in the set and the set is closed. For example, the empty set is closed and the set of integers are closed.

You are right.

Consider this example:

Let $X = \{0 \}$, the set containing just one point: $0$.

Give $X$ the discrete topology $\mathcal{T} = \mathcal{P}(X) = \{ \emptyset, \{0\} \}$.

Then $X$ has no limit points (because there is a neighborhood of $0$ that contains no other points in $X$ but $0$ -- namely, $\{0\}$), but $X$ is certainly closed since the complement of $X$, which is $\emptyset$, is open.