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Let $X$ be a set, then the discrete topology $T$ induced from discrete metric is $P(X)$, which is the power set of $X$

I know $T \subset P(X)$, but how do we know $T=P(X)$

Thank you!

mathshungry
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  • First, note that usually the discrete topology on $X$ is defined as the powerset of $X$ - that is, there's no reference to a metric at all. That said, in this definition: can you show that every subset of $X$ is open? If not, where are you running into difficulty? – Noah Schweber Jan 11 '17 at 17:03
  • If I don't have a metric, how can I define what is open? – mathshungry Jan 11 '17 at 17:07
  • You just define the topology directly. The notion of topology isn't dependent on the notion of metric: a topology on a set $X$ is a collection of subsets of $X$ satisfying some certain properties. $P(X)$ satisfies these properties. Any such collection of subsets is a topology, whether it comes from a metric or not; and indeed there are topologies which cannot come from a metric. Although classes often introduce metrics before topologies, there is no need to do this. – Noah Schweber Jan 11 '17 at 17:23
  • (cont'd) The topology tells you what is open - specifically, the elements of the topology are the open sets. For an example of this, it's not hard to check that $\tau={\emptyset, {a}, {a, b}}$ is a topology on the set $X={a, b}$. In this topology, every subset of $X$ except ${b}$ is open, since every subset of $X$ except ${b}$ is in $\tau$. Note that no metric crops up anywhere in this definition. – Noah Schweber Jan 11 '17 at 17:25

2 Answers2

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For any set $U \in P(X)$ we have that $B_1(x) = \{x\} \subseteq U$ for any $x \in U$. Hence $U$ is open with respect to the topology induced by $d$.

If you are unsure what the metric topology is, you can have a look here. This is a document I am currently working on to understand the connection between topological spaces and metric spaces better myself.

TheGeekGreek
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From the definition of the discrete metric, taking a ball of radius $1/2$ around any element $x \in X$ gives you that $\{x\} \in T$.

Let $A \subset X$ be an element of $P(X)$. Then $$A = \bigcup_{a \in A}\{a\} \in T$$ since any union of elements in $T$ is an element of $T$. This proves that $P(X) \subseteq T$, and you already have $T \subseteq P(X)$, hence $T = P(X)$.

sTertooy
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