$x$-axis is meager set on $\mathbb{R}^2$

Question

Subset $A$ of metric space $X$ is meager on $X$, iff $\text{IntCl}A=\emptyset$.

But, why $x$-axis is meager set on $\mathbb{R}^2$?

My attempt (please don't kill me):

$\text{IntCl}\mathbb{R}=\text{Int}\mathbb{R}=\mathbb{R}\neq \emptyset$

Thank you!

Answer 1

The interior of $\mathbb{R}$ as a subset of $\mathbb{R}^2$ is very different from its interior as a standalone topological space.

Given any point $P$ on the $x$-axis, every open neighbourhood of $P$ contains a point not on the $x$-axis. So $P$ is not in the interior of the $x$-axis. To put it another way, every point of the $x$-axis is on the boundary of the $x$-axis.

Answer 2

Note that the "$x$-axis" as a subspace of $\Bbb R^2$ should be

$$A=\{(x_1,x_2)\in \Bbb R^2:x_2=0\}$$

To aid André's answer:

In mathematics, a nowhere dense set in a topological space is a set whose closure has empty interior [viz ${\rm int}({\rm cl}(A))=\varnothing$].

The order of operations is important. For example, the set of rational numbers, as a subset of $\Bbb R$ has the property that the interior has an empty closure, but it is not nowhere dense; in fact it is dense in $\Bbb R$.

The surrounding space matters: a set $A$ may be nowhere dense when considered as a subspace of a topological space $X$ but not when considered as a subspace of another topological space $Y$. A nowhere dense set is always dense in itself.

Something more, plus terminology:

Every subset of a nowhere dense set is nowhere dense, and the union of finitely many nowhere dense sets is nowhere dense. That is, the nowhere dense sets form an ideal of sets, a suitable notion of negligible set. The union of countably many nowhere dense sets, however, need not be nowhere dense. (...) Instead, such a union is called a meagre set or a set of first category.