30
$\begingroup$

I am self-studying analysis and ran across this:

$\mathbb R \setminus \mathbb N$ is an open subset of $\mathbb R$

My best guess for interpretation was this:

the set $\mathbb R \setminus \mathbb N$ is an open subset of $\mathbb R$.

which doesn't mean much to me. Can anyone clear this up a bit? I know that the 'divided by' symbol is usually a slash in the opposite direction. And I am unsure how I would divide the reals by the naturals anyway.

$\endgroup$
1
  • $\begingroup$ Why use the "\" symbol when "-" is more intuitive and more common? $\endgroup$
    – nickalh
    Jun 5 at 12:47

2 Answers 2

Reset to default
47
$\begingroup$

It’s set theoretic complement and in this case it denotes the set of all reals which are not natural: $$ℝ \setminus ℕ = \{x ∈ ℝ;~x \notin ℕ\}$$

$\endgroup$
3
  • $\begingroup$ makes much more sense! thank you. $\endgroup$
    – 123
    Dec 25, 2014 at 23:31
  • $\begingroup$ @ k.stm - Out of curiosity, it seems I could prove this true by arguing that our set R\N is a union of open subsets and is therefore itself open. Would this, more or less, be correct? $\endgroup$
    – 123
    Dec 26, 2014 at 0:03
  • $\begingroup$ @mathtastic Yes. $\endgroup$
    – k.stm
    Dec 26, 2014 at 0:08
18
$\begingroup$

The backward slash is kind of the set theory equivalent of subtracting, i.e.,

$$A\setminus B=\{a\in{A}\mid a\notin{B}\}\;.$$

$\endgroup$
5
  • 1
    $\begingroup$ Use \setminus to get the set difference operator, and \mid to get the vertical bar with space around it. $\endgroup$
    – Brian M. Scott
    Dec 25, 2014 at 23:32
  • 3
    $\begingroup$ But keep in mind that you still can get something like $\mathbb Q \setminus \{\pi\} = \mathbb Q$. So you just 'substract' as long the thing you want to substract was actually there. $\endgroup$
    – flawr
    Dec 25, 2014 at 23:32
  • 1
    $\begingroup$ Thank you, I was having difficulties with the proper formatting. $\endgroup$
    – Dasherman
    Dec 25, 2014 at 23:33
  • $\begingroup$ thank you for taking the time to provide this answer! $\endgroup$
    – 123
    Dec 25, 2014 at 23:57
  • $\begingroup$ I've seen the following in pseudocode: "A \ B". So this means that x is in Array A, AND not in Array B. The | in your formula doesn't mean OR right? $\endgroup$
    – Vincent
    Mar 14, 2018 at 15:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .