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I'm trying to prove the following statement:

Suppose $\{A_i\}_{i \in \mathbb{N}}$ is a countable, non-increasing collection of non-empty sets, i.e. $A_i \neq \emptyset$ and $A_i \supseteq A_{i+1}$ for any $i \in \mathbb{N}$, then $\bigcap\limits_{i \in \mathbb{N}} A_i \neq \emptyset$.

I suspect the statement to be true, but I am unsure about what kinds of arguments i can use to show it.

In the finite case, a simple induction argument shows that for any $k \in \mathbb{N}$, $\bigcap\limits_{i = 0}^{k} A_i \neq \emptyset$ (actually $\bigcap\limits_{i = 0}^{k} A_i = A_k$). Since this happens for just any $k \in \mathbb{N}$, can I use this as an arguments that it must also hold for the countable intersection (I suspect not)? or do I need some other line of reasoning?

Any pointers to proof methods that can be applied or proofs of other similar statements will be helpful.

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    $\begingroup$ Suppose we take $A_k$ to be all natural numbers greater than or equal to $k$? The sets are nonempty, nested, and their intersection is... $\endgroup$ – hardmath Jan 13 '16 at 11:52
  • $\begingroup$ Consider $A_n = (0, 1/n) \subseteq \Bbb Q$. $\endgroup$ – BrianO Jan 13 '16 at 17:51
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It isn't true. If for example $A_i = \{n\in \mathbb N: n>i\}$ you would have that the intersection is empty.

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  • $\begingroup$ Let me just try and understand the reasoning here. If we take $A_i = \{n \in \mathbb{N} \mid n > i\}$ then $\bigcap\limits_{i \in \mathbb{N}} A_i = \{n \in \mathbb{N} \mid n > m\; \text{for all}\; m \in \mathbb{N}\}$, i.e. the set containing all natural numbers that are strictly larger than all natural numbers, which of course is empty? $\endgroup$ – AcId Jan 13 '16 at 12:04

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