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I have a question that is giving me some minor grief:

If $A$ is a closed set containing all rational numbers $r \in [0, 1]$, then show that $[0, 1] \subset A$.

I don't really understand this question - surely that set of all values $[0, 1]$ contains infinitely more points that the set of rational numbers over the same interval? Am I missing something large?

Thanks in advance.

Edoardo Lanari
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william
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  • Use the fact that any number $x$ is the limit of a sequence of rationals and keep in mind that $A$ is closed. – David Mitra Jul 23 '14 at 10:33
  • What exactly does that mean, the limit of a sequence of rationals? I only just started analysis at university, apologies. – william Jul 23 '14 at 10:36
  • It means given $x$, there is a sequence of rationals $(r_n)$ such that $r_n\rightarrow x$. Have you studied sequences? What characterizations of "closed set'' do you know? – David Mitra Jul 23 '14 at 10:39
  • Do you mean like geometric and arithmetic sequences and so forth? I have studied those, but not exhaustively. – william Jul 23 '14 at 10:42
  • $(r_n)$ is just some sequence. If you haven't studied sequences in general, ignore my previous comment. But, somewhere, you should have been told what a "closed set" is. What is your definition of a closed set? – David Mitra Jul 23 '14 at 10:46
  • Let $R$ be the set of rationals in $[0,1]$ and $A$ be any closed set containing $R$. Note that $R$ is not a closed set. You are required to show that $[0,1]\subseteq A,$ not that $[0,1] \subseteq R.$ The fact that $[0,1]$ contains infinitely many numbers which are not in $R$ is besides the point, because it is $A$ which we are considering. The point is that if you try and define a closed set containing the rationals, you have to include the irrationals in that closed set, too, because the rationals get 'infinitely close' to the irrationals. – Shai Jul 23 '14 at 10:47
  • Is a 'closed set' the same as a 'non open' set? If so then I know it as a set in which the 'non interior' points is not equal to the empty set, etc. So there exists a ball such that you can always find a radius r that is not a subset of the original set. Is any of that correct? – william Jul 23 '14 at 10:49
  • A closed set is not just a non-open set $-$ it is the complement of an open set. – TonyK Jul 23 '14 at 10:59
  • Thank you for clarifying. My course notes are exceedingly bad and have nothing to say of either. ugh – william Jul 23 '14 at 11:00
  • No. A set is closed if its complement is open. With regards to your problem, you need to show that if $x\in[0,1]$, then any ball containing $x$ also contains a rational in $[0,1]$. Alternatively, you need to show $A^C$ is disjoint from $[0,1]$. To do this, pick $x\in [0,1]$. Could $x\in A^C$? If it were, it would be an interior point of $A^C$. What trouble does this cause? – David Mitra Jul 23 '14 at 11:01
  • Thanks for clarifying, I'll give that a go. I have not really done proofs before, despite this being my third semester – william Jul 23 '14 at 11:05
  • "A is a closed set containing all rational numbers" I'm, no sure if this is what confuses you, but: be aware that "a set containing (X)" is not the same as "a set consisting of (X)". – leonbloy Jul 23 '14 at 11:11

4 Answers4

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Since $A$ is closed, it contains all limit points to all sequences as well.

J126
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A is closed, so it contains the closure of the set of all rational points in [0,1], because the closure of a set S is the smallest closed set that contains S. By contrast, the interior of a set S is the greatest open set that is contained in S.

Steven Van Geluwe
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All can be derived from properties of closure operation $\bar{X}$:

$\bar{A} = A$, since A is closed.

Also, if $B~\subset~A$, then $\bar{B}~\subset\bar{A}$.

$\bar{\mathbb{Q}} = \mathbb{R}$, since $\mathbb{Q}$ is dense in $\mathbb{R}$.

Altogether gives you:

. $\mathbb{Q}\cap[0,1] \subset A$ (hypothesis)

. $\overline{\mathbb{Q}\cap [0,1]}~\subset~\bar{A}$

. $[0,1]~\subset~A$

carlosayam
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  • This looks immensely convincing, but I have to ask: what is A bar? Does 'is dense in' mean that every Q is somewhere in R? – william Jul 23 '14 at 10:58
  • $\bar{A}$ is the closure operation: adding all limiting points to a set in the topological sense. Some texts use $Cl(A)$ instead, like in wikipedia: http://en.wikipedia.org/wiki/Closure_(topology). See the link :) – carlosayam Jul 23 '14 at 11:02
  • And a set is dense in other (in topology), when its closure is the whole enclosing set. It is mentioned in wikipedia too. – carlosayam Jul 23 '14 at 11:03
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A closed set contains every point near it. So the question is asking you to show that every point in $[0,1]$ is near the subset of rational numbers between $0$ and $1$.

Maybe it will help to consider the complementary problem: if $A$ is an open set that doesn't contain any rational number between $0$ and $1$, then prove that it is disjoint from $[0,1]$

Or contrapositively, if an open set contains an element in $[0,1]$, then it contains a rational number in $[0,1]$.