I have been asked to describe the following sets, and then prove my answers using the element method, but i am not sure how to do this.

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I am trying to prove that (b) is equal to $0$ as $i$ approaches infinity, and that c is equal to the interval $[0, \frac12]$ as $i$ approaches infinity.

However i have no idea how to do this.

If someone wouldn't mind explaining how I would do this, i would be immensely grateful.

Thanks Tim

  • 1
    $\begingroup$ I think you mean that b equals $[0,1/2]$ and c equals $[0,1)$ $\endgroup$
    – Darth Geek
    May 6 '15 at 11:48
  • 3
    $\begingroup$ Also, there isn't really a limit process a la "as $i$ approaches infinity" involved in the definition of the sets, they are just the intersection/union of infinitely many sets $\endgroup$ May 6 '15 at 11:49
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    $\begingroup$ If any textbook is telling about such a thing as "the element method", the best advice to you is to throw away that book! $\endgroup$
    – user21820
    May 6 '15 at 12:00
  • $\begingroup$ Try inclusion for (c) and (b) (for (b) note that the intervals are closed which gives the answer you mention). $\endgroup$
    – Piquito
    May 6 '15 at 12:13

First you should know the definitions. $\def\nn{\mathbb{N}}$

Given a sequence of sets $(S_i)_{i\in\nn}$, their intersection $\bigcap_{i=1}^\infty S_i$ is defined as $\{ x : \forall i\in\nn ( x \in S_i ) \}$, which is the rigorous way to state "the set of all elements $x$ such that, no matter which natural number $i$ is chosen, $x$ is contained in every $S_i$".

Given a sequence of sets $(S_i)_{i\in\nn}$, their union $\bigcup_{i=1}^\infty S_i$ is defined as $\{ x : \exists i\in\nn ( x \in S_i ) \}$, which is the rigorous way to state "the set of all elements $x$ such that, there is at least one choice of natural number $i$ such that $x$ is contained in $S_i$".

Given two sets $S$ and $T$, the set equality $S = T$ is defined to mean $\forall x ( x \in S \leftrightarrow x \in T )$, which is the rigorous way of saying that no matter what object $x$ you choose, whether it is in $S$ is exactly the same as whether it is in $T$.

Next you should think carefully about what the definitions mean when applied your specific cases, where you want to find a simple form for each set and to prove that it is an equal set to the original. There is no reason or need to use any particular method.

Below I give the thought process to get the solution but please look only after you have intuitively convinced yourself what your sets are and why they are what they are based on the definitions.


Let $S = \bigcap_{i=1}^\infty [0,\frac{i}{i+1}] = \{ x : \forall i\in\mathbb{N} ( x \in [0,\frac{i}{i+1}] ) \}$. What can it be? By definition of equality, if $T$ is an equal set to $S$, then any element that you can find in $S$ must be in $T$. You have named $0$ as one such element. Are there no others? Notice how the sets in the sequence are related to each other. Is it true that one of the sets in the sequence is a subset of each of the others? If so, then clearly that set is the most restrictive in terms of what can be in the intersection. So you should intuitively understand the answer. To prove it rigorously, you need to show that given any element in $S$, it is in that set in the sequence by definition of intersection, and hence in the answer. On the other hand you also need to show that given any element in the answer, it is in every single set in the sequence, for you to be able to conclude that it is in the intersection $S$. That is the harder part but should not be difficult.


Similarly for the union you want the set that contains all the objects that are in at least one set in the sequence. What can it be? This time there is no set in the sequence that is a superset of each of the others, so the answer is not so easy. But you can still intuitively see the answer. Is $0$ an element of the answer? Is $2$? Can you tell me some real number $c$ such that everything below $c$ is in the answer but nothing above $c$ is in the answer? What about $c$ itself? To prove it, again it happens that one part is easy, that is to show that every object that is in some set in the union is also in the answer, which you show by checking every element of each set in the union. The other part is to show that every element in the answer is also an element of some set in the union. This part requires the archimedean property of real numbers.

I hope you can fill in the details by yourself. =)


We are talking here about the increasing sequence of intervals $$J_n:=\left[0, \ {n\over n+1}\right]=\left[0, \ 1-{1\over n+1}\right]\qquad(n\geq1)\ .$$ It is obvious that the intersection of these intervals is the first one, $\bigl[0,{1\over2}\bigr]$. There is no last interval in this sequence (which then would be the union of them); but we can argue as follows: Given any nonnegative number $\alpha <1$ there is an $n$ with ${1\over n+1}<1-\alpha$, or $\alpha<1-{1\over n+1}$. This proves $\alpha\in J_n$, whence such an $\alpha$ is in the union of the $J_n$. On the other hand, none of the $J_n$ contains numbers $\alpha<0$ or numbers $\alpha\geq1$. It follows that $$\bigcup_{n=1}^\infty J_n=[0,1[\ .$$


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