(1) Given an example of sets $A_1\subseteq A_2 \subseteq\dots$ in $\mathbb{R}$ such that each $A_n$ is bounded and $$\bigcap ^{\infty}_{n=1}A_n= \emptyset.$$

(2) Given an example of sets $A_1\subseteq A_2 \subseteq\dots $ in $\mathbb{R}$ such that each $A_n$ is closed and $$\bigcap ^{\infty}_{n=1}A_n= \emptyset.$$ enter image description here

My attempt:

For (2) $A_n=[n,\infty)$

(1) $\left\{\left(0,\frac{1}{n}\right) \right\}^\infty _{n=1}$

Am I correct?

  • 1
    $\begingroup$ You have the containment sign backwards in statement (1). Other than that, it looks good. $\endgroup$ – saulspatz Mar 8 '18 at 3:18
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    $\begingroup$ Your answers do not satisfy $A_k \subseteq A_{k+1}$. The reverse containment is satisfied. Indeed, the stated requirements cannot be satisfied unless $A_1 = \emptyset$, for both problems. $\endgroup$ – Bungo Mar 8 '18 at 3:18
  • $\begingroup$ @Bungo.. you mean that not satisffies increasing property right? $\endgroup$ – Inverse Problem Mar 8 '18 at 3:20
  • $\begingroup$ Correct, your sequences are both decreasing, not increasing. Are you sure the problem statements should not request $A_1 \supseteq A_2 \supseteq \cdots$? $\endgroup$ – Bungo Mar 8 '18 at 3:21
  • $\begingroup$ @Bungo..please can you give me examples $\endgroup$ – Inverse Problem Mar 8 '18 at 3:22

Nope this is incorrect. In fact with $A_1\subseteq A_2 \subseteq..... \subset \mathbb{R}$ we have $$\bigcap ^{\infty}_{n=1}A_n=A_1$$ therefore $$A_1=\emptyset.$$


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