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I'm reading through the Basic Topology of R chapter in Abbott's analysis book just to improve my understanding after studying limits of function and sequence. I can see some connection to basic limit concept and theorem, however I find the discussion pretty thick already in terms of abstractness. I want to easily construct non-trivial example of open and closed set but it's rather difficult.

One of the exercise is this: "Give an example of an infinite collection of nested open sets whose intersection is closed and non empty." (Nested means in this case $A_1 \supset A_2 \supset A_n$)

Is this set correct $A_n=(0,\frac{n+1}{n})$? We have a sequence $\frac{n+1}{n}$ that will converges to $1$. $1$ is always in the intersection of all $A_n$. But $1$ is the limit point for the above sequence, so the intersection is closed and non-empty.

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  • $\begingroup$ This will only "close" one end of your interval; what about "(0"? Your example shows that a half-open interval can be generated by open intervals $\endgroup$ – user128779 Oct 30 '14 at 11:04
  • $\begingroup$ Will my example work if I just set the first end point to be $\frac{1}{n}$ instead of null? $\endgroup$ – vTx Oct 30 '14 at 11:13
  • $\begingroup$ Yes that would work fine since $0 \in (1/n,1 + 1/n)$ for all n > 0 $\endgroup$ – user128779 Oct 31 '14 at 23:49
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$A_n=(-\frac{1}{n},\frac{1}{n})$. The intersection is $\{0\}$, which satisfies the requirement.

Your example is not correct. The intersection of your example is $(0,1]$, which is not closed.

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  • $\begingroup$ So the basic idea is, we want to create a sequence of set (all of them open) such that at large $n$, their intersection actually form a closed set? And the easy way to do it is to set up a sequence at their end points, which converge to one number or perhaps different number? $\endgroup$ – vTx Oct 30 '14 at 11:12
  • $\begingroup$ @vTx Note in the example I given, for any finite intersection of sets are still open. This is true in general, any finite intersection of open sets is still open. It becomes closed only for the infinite intersection. The basic idea is let the end points converge to a number ( or two numbers) which is contained in every interval. $\endgroup$ – John Oct 30 '14 at 11:17
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    $\begingroup$ Ok I get it now. Thanks! $\endgroup$ – vTx Oct 30 '14 at 11:32
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Any neighborhood of 0 will contain points from all your sets An, so the intersection is not closed.

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  • $\begingroup$ For emphasis, any neighborhood of $0$ will contain points belonging to the intersection $\cap A_n$, so the intersection is not empty but also not closed (since $0 \not\in \cap A_n$). $\endgroup$ – hardmath Oct 30 '14 at 11:38

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