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I am having trouble formalizing two proofs I have to make about an infinite intersection of sets.

Suppose that, for every $k\in\Bbb N$ ($k>0$), we define the set $S_k = \{x\in\Bbb R: 0\le x<1/k\}$.

  1. Prove that, for any $k>0$, $0\in S_k$.
  2. Prove that $\{0\}=\bigcap_{k>0}S_k$

For the first one, I am trying to say that, as $x$ is equal or greater than $0$ for $x=0$, and this $x\in\Bbb R$, there is always an $x$ for which that statement is true (say, $x$ being $0$).

The other one is driving me nuts. I do not have any idea on how to make it. I know it is true, because there is always an element $0$ present in all $S_k$, this follows from the reasoning in the first point.

Any help would be greatly appreciated.

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Does my formatting fit with your original intention? –  Cameron Buie Oct 24 '12 at 5:14
    
Yes, thanks a lot. –  Teckizt Oct 24 '12 at 5:15

2 Answers 2

Here is one possible baby step proof for the second part, where we want to prove that $\;\{0\} = \langle \cap k :: S_k \rangle\;$, where I implicitly assume $\;k \in \mathbb N^+\;$.

Let's calculate which elements $\;x\;$ are in the set $\;\langle \cap k :: S_k \rangle\;$ by expanding the definitions and simplifying: \begin{align} & x \in \langle \cap k :: S_k \rangle \\ \equiv & \;\;\;\;\;\text{"definition of $\;\cap\;$ quantification"} \\ & \langle \forall k :: x \in S_k \rangle \\ \equiv & \;\;\;\;\;\text{"definition of $\;S_k\;$"} \\ & \langle \forall k :: x \in \mathbb R \;\land\; 0 \leq x \lt 1/k \rangle \\ \equiv & \;\;\;\;\;\text{"logic: extract parts which don't change with $\;k\;$"} \\ & x \in \mathbb R \;\land\; 0 \leq x \;\land\; \langle \forall k :: x \lt 1/k \rangle \\ \equiv & \;\;\;\;\;\text{"multiply both sides of $\;<\;$ by $\;k\;$, allowed since $\;k > 0\;$"} \\ & x \in \mathbb R \;\land\; 0 \leq x \;\land\; \langle \forall k :: k \times x \lt 1 \rangle \\ \equiv & \;\;\;\;\;\text{"divide both sides of $\;<\;$ by $\;x\;$, with the special case $\;x = 0\;$"} \\ & x \in \mathbb R \;\land\; 0 \leq x \;\land\; \langle \forall k :: x = 0 \lor k \lt 1/x \rangle \\ \equiv & \;\;\;\;\;\text{"logic: extract part which doesn't change with $\;k\;$"} \\ & x \in \mathbb R \;\land\; 0 \leq x \;\land\; (x = 0 \lor \langle \forall k :: k \lt 1/x \rangle) \\ \equiv & \;\;\;\;\;\text{"$\;\forall k\;$ is false for, e.g., $\;k := \lceil 1/x + 1 \rceil\;$"} \\ & x \in \mathbb R \;\land\; 0 \leq x \;\land\; (x = 0 \lor \text{false}) \\ \equiv & \;\;\;\;\;\text{"simplify"} \\ & x = 0 \\ \end{align}

This proves the second part.

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For the first part, all you need is to show that $0<\frac1k$ for all integers $k>0$. For the second part, you must show that if $x>0$, then there is some integer $k>0$ such that $\frac1k\leq x$ (hint: $\Bbb N$ has no upper bound in $\Bbb R$), then the rest of the claim follows from the first part and the definition of the $S_k$'s.

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I don't understand, for the second part, why $1/k≤x$. Wouldn't that be a contradiction? –  Teckizt Oct 24 '12 at 13:58
    
Not a contradiction. That's simply how you show that $\bigcap_{k>0}S_k$ contains no positive numbers. –  Cameron Buie Oct 24 '12 at 17:43

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