# Why are singletons open in a discrete topology?

I was reading through Loring Tu's An Introduction to Manifolds, and I came across this example in the appendix:

For any set $S$, let $\tau$ be the collection of all subsets of $S$. Then $\tau$ is a topology on $S$, called the discrete topology. A singleton set is a set with a single element. The discrete topology can also be characterized as the topology in which every singleton subset $\{p\}$ is open. A topological space having the discrete topology is called a discrete space. The discrete topology is the finest topology on a set.

Right before this example, the author provides a lemma called local criterion for openness, but I am unable to see why a singleton in this situation is open.

EDIT:

Local Criterion for Openness:

Lemma A.2 (Local criterion for openness). Let $S$ be a topological space. A subset $A$ is open in $S$ if and only if for every $p ∈ A$, there is an open set $V$ such that $p ∈ V ⊂ A$.

Proof.

($⇒$) If $A$ is open, we can take $V = A$.

($⇐$) Suppose for every $p ∈ A$ there is an open set $V_p$ such that $p ∈ V_p ⊂ A$. Then $$A \subset \bigcup_{p \in A} V_p \subset A$$ so that equality $A = \bigcup_{p \in A} V_p$ holds. As a union of open sets, $A$ is open.

• The discrete topology is the one in which every subset of $S$ is open; in particular, $\{x\}$ is open for any $x$ Mar 22, 2018 at 6:13
• Yes, but what exactly is the proof of that? Or am I thinking of this wrong? Mar 22, 2018 at 6:15
• Because .... we say it is? What more reason do we need? Mar 22, 2018 at 6:25
• "Or am I thinking of this wrong?" You are. "Open" doesn't have any meaning (except in metric spaces). A set is open iff and only if the set is in $\tau$. And $\tau$ can be any list of subsets (provided intersections and unions are also on the list). The discrete topology simply defines every set is open. That's all there is to it. Mar 22, 2018 at 6:43
• Singletons are always open in the discrete topology... as pointed out, there is nothing to prove... but, you didn't mention what the "local criterion " actually is... if you would include it, we should be able to see how it is met in the case of singletons when the topology is the discrete topology...
– user403337
Mar 22, 2018 at 7:29

In topology the open subsets are nothing more or less then the subsets that we say are open[1]. "Open" by itself doesn't mean anything intrinsic.

Each topology has a list of sets call a "topology" and this list is simply a list of all the subsets that we are going to call open[1]. The discrete topology is one in which the list of all open subset is EVERY subset. So every set is open. Why? Because we said so. We don't need any other reason[1].

[1] Okay, the list must obey a few rules. Any f union of sets on the list must be on the list. And finite intersection of sets must be on the list. And the set itself and the empty set must be on the list. But other than that we can select the list to be anything we want.

We can have a "characterisation" of a topology in which we don't list out all the open sets but just a base few, from which by taking unions and intersections we can determine all the other sets on the list. If we claim that all singletons are open (why? Because we said so.) then it'd follow that all unions of singletons are open. Hence all sets are open.

The discrete topology is a topology in which all sets are open. That is a definition. You can't prove it because there is nothing to prove.

• I know the way now, thanks a lot! Mar 22, 2018 at 6:51

I think you might be overthinking it.

For any set $$S$$, let $$\tau$$ be the collection of all subsets of $$S$$.

It's standard to use the letter $$\tau$$ to denote the set of open sets in a given topological space. In this case, we are told that $$\tau$$ includes every subset of the set $$S$$; this is the very definition of the discrete topology. In particular, singleton sets, i.e. sets of the form $$\{x \}$$ for any $$x \in S$$, are themselves subsets of $$S$$. So $$\{x \} \in \tau$$ for every $$x \in S$$, or in other words, every such set is open.

You may wish to verify this topology indeed satisfies the definition of a topology:

1. $X \in \tau$ and $\emptyset\in \tau$
2. An arbitrary union of elements in $\tau$ is again in $\tau$
3. A finite intersection of elements in $\tau$ is again in $\tau$

So we have simply defined all subsets to be in $\tau$ and call them open since we can. It is perhaps hard to grasp because it is strange to define open sets in this way. It is similar to the first time you see the discrete metric $$d(x,y) = \begin{cases} 0, \;x \ne y \\ 1, \;x=y \end{cases}$$ which doesn't really "seem" like a distance in the intuitive way you know, but satisfies the definition of a metric.