# Open and closed sets of real numbers

Is it true that $\{a\}$ is closed but not open for any real number $a$?

For me, I could first let a set $A=\{a\}$, and let $C=\{\text{collection of all subsets of }A\}$.

Since the definition of topological space requires that both the empty set and $A$ belong to $C$, thus I can conclude that $A$ must be open.

Now, $A$ is closed since its complement, the empty set is open.

Therefore I see it's both open and closed.

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Don't yell, please. (All caps is interpreted as yelling). –  Arturo Magidin Mar 28 '11 at 16:51
Are you the same person as math.stackexchange.com/users/8755/leopold ? You should not create so many aliases! –  Arturo Magidin Mar 28 '11 at 16:53
$A$ as a singleton space is always open and closed, but A is closed and not open in $\mathbb R$ (with the standard topology). –  Alexander Thumm Mar 28 '11 at 16:54
You've been asking this question (or small variations thereof) for several days now, under several different aliases. Please pick one user name and stick to it, and consider editing questions already posed instead of starting entirely new ones. –  Arturo Magidin Mar 28 '11 at 16:58
@Arturo It seems weird that there does not seem to be any sort of reply from the user. Looking back at some of the other questions he has made, he hasn't responded at some queries from other users. In fact the only activity seems to be asking the questions and let them be. Otherwise I don't see a reason to ask over and over minor variations of basically the same question. –  Adrián Barquero Mar 28 '11 at 17:09

"Open" and "closed" are not absolute terms, they are relative terms. A subset of a set is "open" with respect to a particular topology, and "closed" with respect to a particular topology.

The real numbers have lots of topologies you can place on them. However, there is one topology which is used most often, namely, the "standard topology". This is the topology that arises from the metric, which makes a subset $\mathscr{O}\subseteq\mathbb{R}$ open if and only if for every $x\in \mathscr{O}$ there exists $\epsilon\gt 0$ such that $$B(x,\epsilon) = \{r\in\mathbb{R}\mid |r-x|\lt\epsilon\} \subseteq \mathscr{O}.$$

When one asks if a subset of $\mathbb{R}$ is "open" or "closed" and does not specify a topology for $\mathbb{R}$, it is assumed that one is talking about the standard topology of $\mathbb{R}$, the one described above.

So your question is not asking "Can you come up with a topology on $\{a\}$ that makes it open or closed?" (The answer to that question is always "yes", no matter what the set you are looking at is). The question you are asking is really:

Is the set $\{a\}$ closed but not open as a subset of $\mathbb{R}$, in the standard topology?

So, use the definition of "open" for the standard topology to see whether $\mathbb{R}-\{a\}$ is open (that will tell you whether the set is closed in the standard topology); and use the definition of "open" for the standard topology to see whether $\{a\}$ is open.

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