# Topological proof that the interval $[a,b)\subset \mathbb{R}$ is not closed

I want to prove that the interval $[a,b)\subset \mathbb{R}$ is not closed using the definition that a set $A$ in a topological space $X$ is closed iff its complement $X-A$ is open.

Here, the topology on $R$ is the standard order topology.

Munkres in his book "Topology" mentions the claim in passing right after he defines what it means for a set to be closed.

My attempts was:

Assume to the contrary that $[a,b)$ is closed. Then, by definition, its complement $(-\infty,a)\cup[b,+\infty)$ is open. I know that $(-\infty,a)$ is open while $[b,+\infty)$ is closed, but it does not give me a contradiction since the union of an open and a closed need not be not open.

How should I proceed?

I want a topological proof that does not involve epsilon balls and even sequences since the book has not discussed sequences yet.

• How do YOU define open sets on R if not by epsilon balls?
– user223391
Commented May 23, 2015 at 22:50
• If $a \geqslant b$, then $[a,b) = \varnothing$ is closed (and open). So we look at the case $a < b$. Now, if $(-\infty,a) \cup [b,+\infty)$ were open, $b$ would be an interior point of that set. Why is that not so? Commented May 23, 2015 at 22:50
• @avid19 I define open set in $R$ as those equal to the union of the basis sets (a,b) Commented May 23, 2015 at 22:52
• @hl0202 $(a,b)$ is an epsilon ball though!
– user223391
Commented May 23, 2015 at 22:53
• $b$ has no open set around it in your argument...
– DVD
Commented May 23, 2015 at 22:55

The essence of the proofs is the same regardless to what way you choose to define the standard topology on $\Bbb R$.

Recall that an open set is the union of open intervals and open rays. So it suffices to show that $(-\infty,a)\cup[b,+\infty)$ is not the union of open intervals and open rays.

Specifically one point can be exploited here, $b$ itself. Given any open interval $(x,y)$, if $b\in (x,y)$ then $x<b$. If $(-\infty,a)\cup[b,+\infty)$ was open there was an open interval $(x,y)$, or ray when $y$ is $+\infty$, such that $b\in(x,y)$ and $(x,y)\subseteq(-\infty,a)\cup[b,+\infty)$.

But now $x<b$, therefore there is some $z\in(x,b)\cap(a,b)$, but $z\notin(-\infty,a)\cup[b,+\infty)$ which is a contradiction.

• I now understand that $[b,+\infty)$ is not open, but how does it give me a contradiction that $(-\infty,a) \cup [b,+\infty)$ is open? Commented May 23, 2015 at 23:07
• It's closed. It's actually not open. And from this you can derive the complement of $[a, b)$ is not open, so the set is not closed. Commented May 23, 2015 at 23:08
• @hl0202 The proof Asaf gives is conducted on the union, if you observe closely. Commented May 23, 2015 at 23:33
• @GPerez: That's a good point. Commented May 24, 2015 at 6:47

After reading the details of your question, I get the feeling that you (like so many others) mistakenly think that a set can either be open or closed, that somehow open and closed are opposites of each other. I think this is because we use the words open and closed when describing the state of a door -- and a door can't be both open and closed at the same time. (Or maybe it confuses people that a set is closed if its complement -- or in some sense, its "opposite" -- is open, so people may think that the properties are opposite of each other).

BUT in topology we can have a set that is both open and closed. It's actually common and not really mysterious -- once you get past this idea that open and closed are opposites of each other (as they are when describing a door).

To answer your question: Ok, you are on the right track. Assume $[a, b)$ is closed. Then the complement $(-\infty, a) \cup [b, \infty)$ should be open by definition, right? For it to be open, it should be true that for each element in this set, we can find a neighborhood around it entirely contained in the set. Well $b \in (-\infty, a) \cup [b, \infty)$. Now look at any neighborhood around $b$. Is any such neighborhood contained in $(-\infty, a) \cup [b, \infty)$?

The answer is no (you tell me why). Thus, since we found an element in $(-\infty, a) \cup [b, \infty)$ with every open neighborhood around it not contained in $(-\infty, a) \cup [b, \infty)$, then by the definition $(-\infty, a) \cup [b, \infty)$ is not open.