Is the set of irrational number discrete or continuous?

Question

Reading this question and the related answers Is the set of rational number discrete or continuous? I have come up with the question if the set of irrational number, say we call it $\mathbb{I}$, is either discrete or not.

I understood from the answers that there is actually no notion of continuous / discrete for a set, so I'm maybe wrong already in using the terminology.

I also understood that, reading about the rational set, that "it depends on the topology we use". Can I use the same arguments of the accepted answer for proving the set $\mathbb{I}$ is not discrete?

Also, bypassing the wrong terminology, could saying that $\mathbb{Q}$ is not discrete recall the notion of non-open and non-closed in set theory? Like $\mathbb{Q}$ is not discrete but it's not continuous either.

Thank you so much for the patience.

Answer 1

There is a reason we don't use the word "continuous" to describe spaces in mathematics, and it's exactly because of situations like this. The language of topology has more precise terms for describing what's going on here: both the irrational and rational numbers, equipped with their subspace topologies, are

• totally disconnected (so they are nowhere "continuous") but
• have no isolated points (so they are also nowhere discrete, which has a standard and rigorous meaning in topology).

Loosely speaking they can both be thought of as "fractal dust"; points in the dust are not isolated from other points but the points don't "continuously" connect to each other either (e.g. there are no continuous paths from any point to any other point). The Cantor set is also like this.