# “Prove that a topology Ƭ on X is the discrete topology if and only if {x} ∈ Ƭ for all x ∈ X”

This question is from "Introduction to Topology: Pure and Applied," by Colin Adams and Robert Franzosa.

Here's how the authors define a topology:

Let X be a set. A topology Ƭ on X is a collection of subsets of X, each called an open set, such that

(i) ∅ and X are open sets;

(ii) The intersection of finitely many open sets is an open set;

(iii) The union of any collection of open sets is an open set.

Here's how they define a discrete topology:

Let X be a nonempty set and let Ƭ be the collection of all subsets of X. Clearly this is a topology, since unions and intersections of subsets of X are themselves subsets of X and therefore are in the collection Ƭ. We call this the discrete topology on X. This is the largest topology that we can define on X.

Here's where I am with this problem:

First, is this what it's asking? Prove that Ƭ is a discrete topology on X if and only if every x in X is a set {x} in Ƭ?

If that is indeed the question, I'm still struggling. (I'm very bad at proofs).

If we look at the definition of a discrete topology, it seems self evident. A discrete topology must contain all subsets of X, which would include every x in X. I just have no idea how to write a mathematical proof of that.

Appreciate any help.

• You've correctly argued one direction, but you also need the converse. – Randall Jun 4 at 18:00

Proof writing is as obvious as you think it is, just need to write it down formally

($$\implies$$) Assume $$\mathcal{T}$$ is the discrete topology. Then every subset $$S$$ of $$X$$ is open, ie. $$S \in \mathcal{T}$$. This includes all sets containing just a singleton, ie. all sets of the form $$\{ x \}$$ for $$x\in X$$

($$\impliedby$$) Assume $$\{ x \} \in \mathcal{T}$$ for every $$x\in X$$. Then notice that for any subset $$A \subseteq X$$, $$A$$ has (possibly infinite) elements who are members of $$X$$, namely $$A = \{ x_i | x_i \in X \mbox{ and } i\in I \mbox{ where } I \mbox{ is an indexing set} \}$$ But this is exactly a union of singleton sets, $$A = \bigcup_{i\in I} \{ x_i \}$$ which is open as the union of open sets.

Finally, if $$A=\emptyset$$ then it open as the intersection of 2 different open sets, ie. $$\emptyset = \{ x_1 \} \cap \{ x_2 \}$$ where $$x_1 \neq x_2$$.

If $$A=X$$, it is simply the union of all singleton sets of elements in $$X$$ hence open

If $$X$$ has discrete topology, then all subsets are open, specially, for any $$x\in X$$, the singleton $$\{x\}$$ is open.

If all singletons $$\{x\}\subset X$$ are open, then any subset $$A$$ of $$X$$ is open too since $$A=\bigcup_{x\in A}\{x\}$$ is an arbitrary union of open sets.

You correctly argued one direction -- if it has the discrete topology, the topology is the power set of $$X$$, so it contains all subsets of $$X$$, including the singletons $$\{x\}$$.

For the other direction, use axiom (iii) of a topology, and the fact that every subset of $$X$$ can be written as a union of singleton sets $$\{x\}$$.

E.g. $$\{1,2,3 \} = \{1\} \cup \{2\} \cup \{3\}$$.

What topology does $$X$$ have if every subset of $$X$$ is included in it?