Given two non-increasing sequences of sets $$C_k=\{x:2-\frac{1}{k} \lt x\ \leq 2\}$$ $$C_k=\{x:2 \lt x\ \leq 2+\frac{1}{k}\}$$

And defining $$\lim_{k \to \infty} C_k = \bigcap_{k}^{\infty}C_k$$

why is the limit of the first sequence $\{2\}$ and the second $\emptyset$?

I found this answer which explains why the limit of the first sequence is $\{2\}$, The Limit of Nonincreasing Sets: Understanding the Author's Solution, but it does not explain why the second sequence would not work the same way, meaning that both come out to $\{2\}$, as was my first thought.


Consider the first sequence.

Clearly $\{2\} \subset \cap_1^{\infty} C_k$ since $2-\frac{1}{k} < 2 \leq 2$ for all $k$.

Now suppose $x \in \cap_1^{\infty} C_k$. Then $x \in C_k$ for all $k$, so $2 - \frac{1}{k} < x \leq 2$ for all $k$. If $x < 2$, then there exists a $k$ so big that $2-\frac{1}{k} \geq x$, so $x$ would not be in $\cap_1^{\infty} C_k$. Thus we must have $x \geq 2$. If $x > 2$, then it violates the condition of any $C_k$, so then $x$ wouldn't be in $\cap_1^{\infty} C_k$. Thus $x = 2$. It follows that \begin{align*} \cap_1^{\infty} C_k \subset \{2\} \end{align*}

Thus, \begin{align*} \cap_1^{\infty} C_k = \{2\} \end{align*}

In the second sequence, suppose that the intersection $\cap_1^{\infty} C_k$ is not empty. Then there exists an $x \in \cap_1^{\infty}C_k$. Thus $x \in C_k$ for all $k$. This implies that the following statement holds for all $k$, \begin{align*} 2< x \leq 2 + \frac{1}{k} \end{align*} But if $x > 2$, then there exists a $k$ so large that $x > 2+ \frac{1}{k}$. For this particular $k$, we have $x \notin C_k$. Thus $x \notin \cap_1^{\infty} C_k$, a contradiction.

It follows that $\cap_1^{\infty} C_k = \varnothing$ in this case.

  • $\begingroup$ I'm not the OP, but in your first paragraph, don't you have to show first that $\cap_{1}^{\infty}C_k$ is non-empty to be able to conclude that $x = 2$? $\endgroup$ – Pel de Pinda Jun 28 '17 at 18:09
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    $\begingroup$ $\{2\}$ is a subset of the intersection, so it can't be empty. see my edits - i just rearranged some of the sentences $\endgroup$ – Daniel Xiang Jun 28 '17 at 18:12
  • $\begingroup$ Can you explain further the bit about the contradiction? In particular "then there exists a $k$ so large that $x > 2 + \frac{1}{k}$" is confusing me, as I can't visualize why this must be true. $\endgroup$ – rocksNwaves Jun 28 '17 at 18:16
  • $\begingroup$ If $x > 2$, then there exists $\epsilon > 0$ such that $x > 2 + \epsilon$. Choose $k$ so large that $\epsilon > \frac{1}{k}$. Then $x > 2 + \frac{1}{k}$. $\endgroup$ – Daniel Xiang Jun 28 '17 at 18:28
  • $\begingroup$ @DanielXiang so essentially, no matter what $x$ you pick, you can always find a $k$ large enough to put that $x$ outside of the bounds? $\endgroup$ – rocksNwaves Jun 29 '17 at 13:31

In the first case, $2 \in C_k$ for all $k$, hence also to $\bigcap C_k$. And there are no other $x$ which belongs to all $C_k$, hence $\bigcap C_k=\{2\}$.

In the second case, there is no $x$ such that $x \in C_k$ for all $k$. Hence $\bigcap C_k=\emptyset$.


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