If $I_a$ is a sequence of closed/bounded intervals and I am supposed to show using the NIT that the intersection is not equal to the empty set how do I know that $I_1\supseteq I_2\supset \cdots\supset I_n$ for $n\in \mathbb{N}$ and not that $I_n\subseteq\cdots\subseteq I_2\subseteq I_1$? for example consider the set $A:=[\frac{1}{n}, 2]$ where $A_1=[1,2]$ and $\lim_{n\to\infty} A_n=[0,2]$

Edit: Sorry the 2nd one was supposed to be $I_n\supseteq I_{n-1}\supseteq \cdots \supseteq I_1$ as you can see from my example $A_n>A_1$

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    $\begingroup$ It's almost the same thing - except that you're using proper superset in the first chain and non-proper subsets in the second. $\endgroup$ – skyking Oct 23 '15 at 6:15
  • $\begingroup$ If you could look it over once more I accidentally messed up the question. $\endgroup$ – Craig Oct 23 '15 at 6:27

You don't. $A\subseteq B$ and $B\supseteq A$ means the same thing. The same thing goes for proper subset and supersets. The first chain means:

$$I_n\subset...\subset I_2\subseteq I_1$$

while the second means

$$I_n\subseteq...\subseteq I_2\subseteq I_1$$

The first implies the second so you can't have the first to be true without having the second.

If you meant it the second the other way around you would now that they're contradicting. If $A\subset B$ then $A\subseteq B$ is definitely false.

If you meant using non-proper subsets all the way it becomes somewhat interresting.

So for a concrete case where $I_n = [1/n, 2]$ you can easily see that $I_j\supset I_k$ if $j>k$ because the implication that $x\in I_k$ implies that $x\in I_j$ in combination that you can find $x\in I_j$ that's not in $I_k$ (the later makes the superset proper). You only have to consider the definitions for the internvals to see this (especially that $1/k\le x$ implies that $1/j\le x$ if $j>k$ and for being proper you can consider $1/j\le x=1/j$ but $1/k> x=1/j$).

You could also apply the nested interval theorem too, which implies that $I_j\supset I_k$ if $j>k$ immediately (bot not the converse). The proof above is basically similar to the proof of the nested interval theorem.

Note that you could have the both ways, but that means that all $I_j$ are equal (that's why you need to make them non-proper).

  • $\begingroup$ Sorry I wrote the question incorrectly it's supposed to be for the second one that $I_n\supseteq I_{n-1}\supseteq \cdots \supseteq I_1$ $\endgroup$ – Craig Oct 23 '15 at 6:22
  • $\begingroup$ Okay this all makes sense but is it still possible for me to apply the nested interval theorem then? Unfortunately I don't know what my set is defined to be just that it is a closed/bounded interval. How would I go about applying the nested interval theorem then since it is written in my textbook as $I_1\supseteq I_2 \supset\cdots \supset I_n$ and they don't mention anything about cases like $A=[1/n,2]$? $\endgroup$ – Craig Oct 23 '15 at 6:51
  • $\begingroup$ @Craig Yes it applies - I just didn't reckon someone actually named this a theorem. Actually you could come up with a range of variants of it, in this case it's used the addition that both the endpoints don't get to be the same in which case you actually will have the case of proper supersets. $\endgroup$ – skyking Oct 23 '15 at 7:22

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