# Are singletons closed or open in the lower limit topology?

Recall $(\mathbb{R}, \mathcal{T}_\text{lower limit})$ where lower limit topology $\mathcal{T}_\text{lower limit} = \mathcal{T_\mathcal{B}}$ where $\mathcal{B} = \{[a,b) \subseteq \mathbb{R}, a < b\}$

Question: Are singleton sets $\{a\}, a \in \mathbb{R}$ open?

Attempt:

$\{a\}^c = \mathbb{R}\backslash\{a\} = (-\infty, a) \cup (a, \infty) \in \mathcal{T}_{lowerlimit}$

So $\{a\}$ is closed

Correct?

• Question: asks you whether they are open or not. A set can be both open and closed,clopen , in a topology, e.g. $\mathbb{R},\emptyset$. – Emre Jun 5 '16 at 2:33
• @Emre how to show that the singletons are open? – Carlos - the Mongoose - Danger Jun 5 '16 at 2:37

You are correct that singletons are closed, but it might be worthwhile to show why exactly $(-\infty, a)$ and $(a, \infty)$ are open for each $a \in \mathbb{R}$. Use sets of the form $[-n, a)$ for $(-\infty, a)$ and something similar for the other. Note that singletons can't be open in this topology. If they were, then what could you say about any arbitrary subset $A \subseteq \mathbb{R}$?
• Then any arbitrary subset is open because $A = \cup_{x \in A} \{x\}$ – Carlos - the Mongoose - Danger Jun 5 '16 at 2:40
• @Lookbehindyou Exactly. Now find a subset of $\mathbb{R}$ that's not open in $(\mathbb{R}, \mathcal{T}_{lower\; limit})$. – Mnifldz Jun 5 '16 at 2:41
• @Emre $\mathbb{Q}$. – Mnifldz Jun 5 '16 at 5:14
• Showing something is closed means nothing at all (consider $[a,b)$ in the lower limit topology). And it's true that not all singletons can be open at the same time, but the question is, "is there some open singleton?" – Henno Brandsma Jun 5 '16 at 6:00
It's true that singletons are closed. But singletons could also be open. You haven't shown. It is true that if all singletons are open, the space is discrete, which it is not. But there still could be some freak singleton that is open, maybe? Showing something is closed does not mean it's not open: in the lower limit topolog $[0,1)$ is both open and closed! So try a direct proof, not using closedness:
Suppose $\{p\}$ is open. Then there must be a basic open set of the lower limit topology, i.e. a set of the form $[a,b)$, such that $p \in [a,b) \subseteq \{p\}$. In other words, all points in this $[a,b)$ must be equal to $p$. But this is clearly absurd as $[a,b)$ is infinite (or just notice that both $a$ and $a + \frac{b-a}{2}$ are in it). So $\{p\}$ cannot be open. And this holds for any $p \in \mathbb{R}$.