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Hello I was doing a practice problem and I came across something that kind of tripped me up. The question says:

Suppose the Collection $F$ is given by $F=\{ [1, 1+ 1/n] : n\in \mathbb{N}\}$

Find the union of this collection of sets and find the intersection of these sets (I'm struggling to type this).

I looked at the solution and I don't understand why the intersection is $\{1\}$. This element is not actually contained in any of the sets since all the elements of the set are intervals. Can anyone please explain why this is? Intuitively on some level it makes sense since $1$ is the only common "thing" for lack a of a better, word but isn't the intersection of a collection of sets be the elements that are common to all the sets?

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If $\cal A$ is a collection of sets then we define the two operations:

  1. The union of $\cal A$ which is $\bigcup{\cal A}=\{x\mid\exists A\in{\cal A}: x\in A\}$. So the elements of the union are all those which are elements of elements of the collection $\cal A$.
  2. The intersection of $\cal A$ which is $\bigcap{\cal A}=\{x\mid\forall A\in{\cal A}:x\in A\}$. Similarly, the elements of the intersection are those which are elements in all the elements of $\cal A$.

So now we look for $\bigcup F$ and $\bigcap F$. What real numbers are in all the intervals? How can you describe that set nicely? What is the collection of real numbers which appear in some of the sets?

Let me hint you for the first answer, $1$ is clearly in all the intervals and no number smaller than $1$ is in any interval to begin with. For every number $x>1$ we can find an interval $[1,1+\frac1n]$ such that $x$ is not in that interval. So which numbers appear in all the intervals?

To your final paragraph, let me add that $\{1\}$ is in fact the interval $[1,1]$. And it goes to show you that the intersection of non-degenerate intervals doesn't have to be a non-degenerate interval.

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  • $\begingroup$ Thank you, you're notation of union and intersection which defines them as set notation is incredibly helpful. $\endgroup$ – Roger Tynes May 14 '13 at 19:43
  • $\begingroup$ Roger, this is a common notation... I can't possibly take the credit to myself. :-) $\endgroup$ – Asaf Karagila May 14 '13 at 19:48
  • $\begingroup$ Still it's very useful, and for some reason not included the textbook we use. The elements of the elements is basically the key breakthrough that helps me understand this notation; I just discovered this site and it's so useful! $\endgroup$ – Roger Tynes May 14 '13 at 19:58
  • $\begingroup$ I'm glad that you find it helpful, and I'm glad that I could be a part of that as well. $\endgroup$ – Asaf Karagila May 14 '13 at 19:59
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The element $1$ is contained in all of the sets of the collection; the set $\{1\}$ is not a member of the collection, but it is a subset of each member of the collection. (It actually is an interval, by the way: $\{1\}=[1,1]$, a degenerate closed interval. By definition $[a,b]=\{x:a\le x\le b\}$, and if $a=b$ that set contains just the one element $a=b$.)

For each $n\in\Bbb Z^+$ let $F_n=\left[1,1+\frac1n\right]$, and let $\mathscr{F}=\{F_n:n\in\Bbb Z^+\}$. ($\mathscr{F}$ is the collection called $F$ in your question; I’ve changed the name to $\mathscr{F}$ to distinguish it more clearly from the names that I’ve given to the individual sets in the collection.) You want

$$\begin{align*} \bigcap\mathscr{F}&=\bigcap_{n\in\Bbb Z^+}F_n\\\\ &=\bigcap_{n\ge 1}\left[1,1+\frac1n\right]\\\\ &=[1,2]\cap\left[1,\frac32\right]\cap\left[1,\frac43\right]\cap\left[1,\frac54\right]\cap\ldots\cap\left[1,\frac{n+1}n\right]\cap\ldots\;. \end{align*}$$

This is the intersection of a bunch of sets of real numbers, so it will itself be some set of real numbers. Specifically, it’s the set of all real numbers that are in all of the sets $F_n$:

$$\begin{align*}\bigcap\mathscr{F}&=\{x\in\Bbb R:x\in F_n\text{ for all }n\in\Bbb Z^+\}\\\\ &=\left\{x\in\Bbb R:1\le x\le 1+\frac1n\text{ for all }n\ge 1\right\}\;. \end{align*}$$

If $n$ is any positive integer, it’s certainly true that $1\le 1\le 1+\frac1n$, so $1\in F_n$. Thus, $1\in\bigcap\mathscr{F}$. Suppose that $x\ne 1$. If $x<1$, then obviously $x$ does not belong to any of the intervals $F_n$. If $x>1$, then $x-1>0$, and there is a positive integer $n$ large enough so that $\frac1n<x-1$. But then $x>1+\frac1n$, so $x\notin\left[1,1+\frac1n\right]=F_n$. That is, if $x$ is any real number other than $1$, we can find an $n\in\Bbb Z^+$ such that $x\notin F_n$, and therefore $x\notin\bigcap\mathscr{F}$. Thus, $1$ is the only real number that belongs to all of the intervals $F_n$, so it’s the only thing in the intersection of all those intervals:

$$\bigcap\mathscr{F}=\{1\}\;.$$

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  • $\begingroup$ Thank you! I have a much better intuitive sense of it now, I'm still a little bit confused by the notation of union/intersections of families of sets but I've been reading some more stuff on this site and it's starting to make more sense. Thanks! $\endgroup$ – Roger Tynes May 14 '13 at 19:42
  • $\begingroup$ @Roger: You’re welcome! I deliberately used a variety of notations in hopes that it would help. $\endgroup$ – Brian M. Scott May 14 '13 at 19:43
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First note that $1$ is an element of each of the intervals of the form $[1,1+\frac{1}{n}]$, and thus $1$ is in the intersection. Now, assume that $x$ is in the intersection. Then $x$ is in all the mentioned intervals, thus $1\le x\le x+\frac{1}{n}$ holds for all $n\ge 1$. That is only possible if $x=1$, this shows thus that the only element in the intersection is $1$. In other words: the intersection is the set $\{1\}$.

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  • $\begingroup$ Thank you very much for your quick response, I read it last night but didn't get a chance to respond, I'm beginning to understand it now. $\endgroup$ – Roger Tynes May 14 '13 at 19:43
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The interval $[1,1+1/n]$ is a closed interval that contains both endpoints. Normally an open interval is denoted as $(a,b)$, although some people use $]a,b[$.

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  • $\begingroup$ Yeah I always forget to look at it as an interval, I saw it as a set and I forget that intervals are sets. $\endgroup$ – Roger Tynes May 14 '13 at 19:51

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