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T/F: Let $A_n$ = $(0, \frac{1}{n})$ (a bounded set), and $B_n$ = [$n, \infty)$ (a closed set). Any finite intersection of $A_n$'s and $B_n$'s is non-empty, but the infinite intersection of $A_n$'s and $B_n$'s is empty.

I think the answer is true, but my logic behind it is a bit iffy. Does it involve limits? This is my reasoning: As $n \rightarrow k$, where $k$ is some finite number, then $(0, \frac{1}{k})$ and $B_n$ = [$k, \infty)$ have finitely many numbers in it. So, we have a finite intersection of finite numbers, which makes it non-empty. On the other hand, As $n \rightarrow \infty$, then $(0, \frac{1}{\infty})$ and $B_n$ = [$\infty, \infty)$, then they're approaching "emptiness"? I can't really grasp the ideas behind these finite/infinite intersections, so any clarification would be tremendously helpful. Thank you.

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  • $\begingroup$ I suggest you forget about limits for the moment. $\endgroup$ – Lord Shark the Unknown Sep 14 '17 at 5:48
  • $\begingroup$ What should I be using/thinking of to understand this problem? $\endgroup$ – Max Sep 14 '17 at 5:55
  • $\begingroup$ $(0,1/k)$ and $(k,\infty)$ do not "have finitely many numbers in it." Both are infinite sets. Note that if $n_1 < n_2 < \cdots < n_k$, then $A_{n_1} \cap A_{n_2} \cap \cdots \cap A_{n_k} = A_{n_k}$, and $B_{n_1} \cap B_{n_2} \cap \cdots \cap B_{n_k} = B_{n_k}$, so both intersections are nonempty. Why does this argument fail if we intersect infinitely many $A$'s or $B$'s? $\endgroup$ – Bungo Sep 14 '17 at 5:55
  • $\begingroup$ Does it involve countability/uncountability arguments? $\endgroup$ – Max Sep 14 '17 at 5:56
  • $\begingroup$ P.S. Both parentheticals are incorrect. $A_n = (0,1/n)$ is not a closed set, and $B_n = [n, \infty)$ is not a bounded set (unless you're using something other than the usual topology on $\mathbb R$). In any case, topological concepts are irrelevant to the question being asked. $\endgroup$ – Bungo Sep 14 '17 at 6:05
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True

  1. Let $x \in \bigcap\limits_n A_n$, then $x > 0$, but $\mathbb{Q}$ is dense in $\mathbb{R}$, hence there exist rational $0 < \frac{p}{q} < x$ and obviously $\frac{1}{q} \leq \frac{p}{q}$, i.e. $x \notin A_p$ and hence $x \notin \bigcap\limits_n A_n$. Contradiction. Hence, $\bigcap\limits_n A_n = \emptyset$.
  2. Let $x \in \bigcap\limits_n B_n$. Trivially, $x < \lceil{x}\rceil + 1$, hence $x \notin B_{\lceil{x}\rceil + 1}$ and then not in $\bigcap\limits_n B_n$. Same conclusion follows.
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