7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Do you agree that $(-\infty, a)$ is open? $\endgroup$ – Zhanxiong Aug 19 '15 at 2:22
  • $\begingroup$ Yes, but I don't want an answer to be that the complement of an open set is closed. $\endgroup$ – bissi Aug 19 '15 at 2:24
  • 1
    $\begingroup$ $\infty$ is not a limit point of $[a,\infty)$ in $\mathbb R$! $\endgroup$ – user251257 Aug 19 '15 at 2:29
  • 2
    $\begingroup$ $\infty$ is not a limit point. Your set lives in $\mathbb{R}$. A limit point must also belong to the same space. $\endgroup$ – Pburg Aug 19 '15 at 2:29
  • 1
    $\begingroup$ 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. $\endgroup$ – Akiva Weinberger Aug 19 '15 at 4:28
11
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Beat me by six sectiods, drat. :-) $\endgroup$ – Mike Haskel Aug 19 '15 at 2:34
  • $\begingroup$ @MikeHaskel Yet you still deserve an upvote. :) $\endgroup$ – layman Aug 19 '15 at 2:34
5
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ Fixed the error @user251257 pointed out. $\endgroup$ – Mike Haskel Aug 19 '15 at 2:49
  • $\begingroup$ Thanks this is helpful too. Unfortunately I could only mark one answer correct, and the other one provided me with more details I needed. $\endgroup$ – bissi Aug 19 '15 at 3:15
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

How about this? A closed set is a set such that it's complement is open.

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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