# Simple question on closed sets

A closed set is one which contains all its limit points. Why is $[a, \infty)$ closed? Specifically I don't understand how $\infty$ which is a limit point, but it is not in the set.

• Do you agree that $(-\infty, a)$ is open? Aug 19, 2015 at 2:22
• Yes, but I don't want an answer to be that the complement of an open set is closed. Aug 19, 2015 at 2:24
• $\infty$ is not a limit point of $[a,\infty)$ in $\mathbb R$! Aug 19, 2015 at 2:29
• $\infty$ is not a limit point. Your set lives in $\mathbb{R}$. A limit point must also belong to the same space. Aug 19, 2015 at 2:29
• Since "closedness" is relative to the topology, you might wonder if there's a related notion that doesn't depend on the topology. There is: compactness. Aug 19, 2015 at 4:28

The question of why $[a, \infty)$ is closed even though the limit point $\infty$ isn't in it is actually a really good question. Here is the answer:

It depends on what the topological space is. If our space is $\Bbb R$ (i.e., $(-\infty, \infty)$) with its usual topology, then since $\infty$ is not in $\Bbb R$, it's not a limit point of $[a, \infty)$.

Now, if our topological space is the extended reals $\Bbb R^{*}$, which is defined as $[-\infty, \infty]$ with topology generated by all usual open intervals in $\Bbb R$ but also all intervals of the form $(a, \infty]$ and $[-\infty, b)$, then $\infty$ is a limit point of $[a,\infty)$, so $[a,\infty)$ is not closed in $\Bbb R^{*}$.

So, a set being closed is a relative property. A set can be closed in one topological space, but not in another, as we just saw. As a subset of $\Bbb R$, $[a, \infty)$ is closed, but as a subset of $\Bbb R^{*}$, it's not. So when you talk about openness and closedness, you really need to be aware of which topological space you are talking about first.

• Beat me by six sectiods, drat. :-)
– user231101
Aug 19, 2015 at 2:34
• @MikeHaskel Yet you still deserve an upvote. :) Aug 19, 2015 at 2:34

The key point is that whether a set is open or closed depends on what space we're talking about. $[a,\infty)$ isn't "closed" by itself, it's closed in $\mathbb{R}$. There is no point "$\infty$" in $\mathbb{R}$, so it doesn't count for deciding whether or not $[a,\infty)$ is closed in $\mathbb{R}$. In contrast, sometimes we do talk about "$\overline{\mathbb{R}}$", which is a notational convention for $\mathbb{R}$ together with the extra points $\pm \infty$. In $\overline{\mathbb{R}}$, the set $[a,\infty)$ is not closed.

(Don't confuse the $\overline{\mathbb{R}}$ notation with the notation for the closure of a set; they're different concepts.)

• Fixed the error @user251257 pointed out.
– user231101
Aug 19, 2015 at 2:49
• Thanks this is helpful too. Unfortunately I could only mark one answer correct, and the other one provided me with more details I needed. Aug 19, 2015 at 3:15

Take $x\notin [a,\infty)$. It must be that $x<a$. Then consider any open ball of radius less than $\vert a-x \vert$. This ball will not contain any points in $[a,\infty)$. Hence, $x$ is not a limit point by the definition of limit point, because every ball around a limit point must contain another element of the set. Therefore, every limit point of $[a,\infty)$ is also contained in this set.

For $[a,\infty)$ the complement is $(-\infty,a)$ with a point at infinity. This set is clearly open since adding an arbitrary epsilon to infinity yields infinity. Therefore the set $[a,\infty)$ is closed.
But, if the space you're in considers $(-\infty,-\infty)$ open, then you have a topology leading the definitions. Indeed, infinity will be considered a limit point and then $[a,\infty)$ is actually open!