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

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!

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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.

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But does this show the closure of R as a subset of $R^2$ is empty? In other words, I am asking what IS the closure of $R$ as a subset – Tyler Hilton Dec 15 '13 at 5:11
The $x$-axis is a closed subset of $\mathbb{R}^2$. – André Nicolas Dec 15 '13 at 6:43

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

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

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.