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

Question

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.

Answer 1

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

Answer 2

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?

Answer 3

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.

Answer 4

If $(\mathbb X,\mathbb T)$ is a discrete topology $\implies$ every subset of $\mathbb X \in \mathbb T$, hence $ \{x\} \in \mathbb T \forall x \in X$.

Conversely, if $ \{x\} \in \mathbb T \forall x \in \mathbb X$, then for any non-empty set $\mathbb A \subset X$, we have $$ \mathbb A = \cup {x} \forall x \in \mathbb A$$
This implies $\mathbb A \in \mathbb T$.