# Question about discrete topology of $\mathbb{R}$

If we let $$\tau$$ be a topology of the $$\mathbb{R}$$ s.t $$\tau = \mathbb{P}(\mathbb{R})$$, surely $$\{1\} \in \tau$$ and that is an open set if we define such to be an element of a topology. I'm not understanding how this is consistent with other definitions of open set such as

A set $$U \subset R$$ is called open, if for each $$x \in U$$ there exists an $$\epsilon > 0$$ such that the interval $$(x - \epsilon, x + \epsilon)$$ is contained in $$U$$.

In this case, there is no $$\epsilon > 0$$ s.t $$(1 - \epsilon, 1 + \epsilon) \in [1]$$, so how can it be an open set? Does $$\{1\} \neq [1]$$? If it does, how is $$[1]$$ an open set if it's closed off by brackets?

Not all spaces are metric, and thus have open sets characterized by epsilon balls. For instance the trivial topology: $$\{\emptyset,X\}$$.

However discrete spaces are.

The discrete metric is $$d(x,y)=\begin {cases}1,\,x\ne y\\0,\,x=y\end {cases}$$.

Thus each point is open, as $$x=B(x,1)$$, say.

• The interval $(x-1,x+1)$ is NOT the same as the unit ball in the discrete metric. The interval depends on the ordering, NOT on the metric. The ball, on the other hand, very much does depend on the metric. Intervals only make sense when you have an ordered set, in which case $(x-1,x+1)=\{z\mid x-1<z<x+1\}$ which is very much NOT {x}. You've confused one circumstance where the objects coincide with a circumstance where they do not. Sep 17, 2022 at 20:27
• Hmm. I see. @Aaron. Thanks for the touch up. Sep 17, 2022 at 20:38
• @Aaron: Can you think of a case where the two don't coincide?
– MSIS
Sep 17, 2022 at 20:53
• @MSIS Yes. Any time you put a metric on $\mathbb R$ that is not the standard metric, you get that there is an $x, r$ such that $B(x,r)\neq (x-r,x+r)$. The two notions of balls and symmetric intervals like this ONLY coincide for the standard metric. Sep 17, 2022 at 20:56
• Thanks, Aaron. Is the order topology on the Real line equivalent to that generated by the standard metric d(x,y)=|x-y|? I guess any relationship breaks down for higher dimensions, as there's no " obvious" ordering in $\mathbb R^2$ or higher?
– MSIS
Sep 18, 2022 at 19:14

The second definition is for the standard topology of $$\mathbb R$$, not the discrete topology. In the standard topology, the intervals form a "basis" for the topology. If you have a basis $$\mathcal B$$, then a set $$U$$ is open if for every $$x\in U$$ we can find $$V\in B$$ such that $$x\in V\subset U$$.

In the discrete topology, one can take the set of all singletons to be a basis. Then if $$x\in U$$, we can take $$x\in \{x\}\subset U$$. Singletons are replacing intervals.

But again, this is two separate topologies, so the fact that they yield different open sets is what you would expect (unless the two topologies unexpectedly coincided, which they do not here).