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Let $S$ be an infinte subset of $\mathbb{R}$ such that intersection of $S$ with $\mathbb{Q}$ is empty. Then which of following statements is true?

  1. $S$ must have a limit point which belongs to $\mathbb{Q}$.
  2. $S$ must have a limit point which belongs to $\mathbb{R}\backslash \mathbb{Q}$.
  3. $S$ cannot be a closed set in $\mathbb{R}$.
  4. $\mathbb{R}\backslash S$ must have a limit point which belongs to $S$.

I took $S$ as the set of irrational numbers, so the set of limit points of $S$ is $\mathbb{R}$. Now, $\mathbb{R}$ is not contain in $\mathbb{Q}$, therefore the 1st claim is incorrect. $\mathbb{R}\backslash\mathbb{Q}$ has elements of form $\{a+Q : a\text{ is irrational number}\}$. This implies that $\mathbb{R}$ is not contain in $\mathbb{R}\backslash\mathbb{Q}$. Therefore the 2nd option is incorrect as well. Now, the chosen $S$ is not a closed set in $\mathbb{R}$. Hence 3rd option is correct. What will be the limit point of $\mathbb{R}\backslash S$ with the chosen $S$. How can I contradict the 4th claim?

And am I proceeding correctly?

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    Did you notice, when you were typing the post, that the choices have all merged together into an unnecessarily hard-to-read paragraph? You'll need to double space for line breaks. – 2017-01-24
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    Let S = {n+pi| natural}. Statements 1-3 are not true for S. 4) Q is a subset of R/S. Q has limit points at any irrational so any s in S is a limit point of R/S. – 2017-01-24

3 Answers 3

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No, you're not reasoning correctly. In fact, everything you've said wrong.

1 and 2 are about the existence of a single such point, not the assertion that all points satisfy the property. Indeed, when $S=\mathbb{R}\setminus\mathbb{Q}$ then $S$ has $2$ as a limit point and $\sqrt{2}$ so both 1 and 2 hold. Your reasoning for 3 is also wrong, as you have given one example but not proven that it's true for all $S$.

Keeping these issues in mind, try again. Mouse over the spoiler to see the proofs.

Considering $S=\{\sqrt{2}+k:k\in\mathbb{N}\}$ shows that $S$ doesn't have to have any limit points, making both 1 and 2 false. This set is closed, so 3 is also false. 4 is true, because $\mathbb{Q}\subseteq\mathbb{R}\setminus S$. Notice that $S\subset\mathbb{R}=\overline{\mathbb{Q}}$, so every point of $S$ is a limit point of $\mathbb{R}\setminus S$ (in fact, every point of $\mathbb{R}$ is a limit point)

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    Not a fan of the famous proof-by-example method? :D – 2017-01-24
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    @Mathematician42 only when it's a $\Sigma^0_1$ statement. – 2017-01-24
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First, note that your disprove of the first and second statement are incorrect. The claims are not about all the limit points of $S$, but just about the existence of some limit point satisfying the property, i.e., if there is just one limit point of $S$ satisfying the claim, the claim holds. In your case, the set of limit points of the set of irrational number is $\mathbb{R}$, but $\mathbb{R}$ contains both points that belongs to $\mathbb{Q}$ and points that belongs to $\mathbb{R}\backslash S$. Thus, for your chosen $S$, there are both limit points of $S$ belonging to $\mathbb{Q}$ and $\mathbb{R}\backslash S$.

Second, as I notice a wrong use of language in the edit, I will point at it. When we have two subsets $X$ and $Y$ such that $X\subseteq Y$, we don't say "$X$ does not belong to $Y$", but "$X$ is contained in $Y$". The expression "$x$ belongs to $X$" refers to elements inside a set, i.e. $x\in X$. This distinction between a set and its elements is important and it seems to be fuzzy in your arguments, as you seem to interpret $\mathbb{R}\not\subseteq \mathbb{Q}$ as no element of $\mathbb{R}$ belongs to $\mathbb{Q}$ which is obviously false as $\mathbb{R}$ contains lots of elements which belong to $\mathbb{Q}$.

Third, consider $S_0$ the set of irrational numbers and $S_1$ the set of integers plus $\sqrt{2}$ (i.e. $\sqrt{2}+\mathbb{Z}:=\{\sqrt{2}+n\,|\,n\in\mathbb{Z}\}$). By the above claims, you can see that for $S_0$ the only claim not holding is the 3rd claim. However, $S_1$ is closed, so the 3rd claim is false for it. Further, one can show that $S_1$ has no limit points at all (prove it!), so the $1$st and $2$nd claim are also false for it. Hence the only possible true claim is the fourth one (assuming the question is well-posed).

Fourth, in order to prove the 4th claim, just note that $\mathbb{R}\backslash S$ contains $\mathbb{Q}$ and so every accumulation point of $\mathbb{Q}$ is an accumulation point of $\mathbb{R}\backslash S$. Now, every real number is an accumulation point of $\mathbb{Q}$, so, in particular, every point of $S$ is an accumulation point of $\mathbb{Q}$. Hence $\mathbb{R}\backslash S$ must have accumulation points belonging to $S$.

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(4) is true: $\mathbb R$ \ $S\supset Q$ and every real is a limit point of $\mathbb Q$ so every real is a limit point of $\mathbb R$ \ $S$.

To see how $S$ may be closed, let $\mathbb Q=\{q_n: n\in \mathbb N\}$ and let $$\mathbb R \backslash S=\cup_{n\in \mathbb N}(-2^{-n}+q_n, 2^{-n}+q_n).$$ Then $R$ \ $S$ is open so $S$ is closed. The sum of the lengths of the intervals comprising $\mathbb R$ \ $S$ is only $1,$ so an infinite amount of $\mathbb R$ must be in $S.$

Your reasoning on (1) and (2) is wrong. The assertion "$S$ must have a limit point in $T$" (for any given $T$) does NOT mean "$S$ must have a limit point, and all its limit points belong to $T$".

However $S$ may fail to have any limit points at all, for example, if $S=\{n+\sqrt 2: n\in \mathbb N\}.$