Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

2 Answers

up vote 10 down vote accepted

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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.