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I saw an example in a somewhat random and old math book (it was full of exercises, not theory):

$$\bigcup_{n = 1}^{\infty} {\left[ {0 + \frac{1}{{n + 1}};1 - \frac{1}{{n + 1}}} \right]} = \left( {0;1} \right)$$ (union of intervals)

Are there reasonable arguments as to why

$$\left[ {0 + \mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + 1}};1 - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + 1}}} \right] = \left[ {0;1} \right]$$

is not a member of the union? Or maybe there is a chance it depends on particular literature..?

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I think you mean "is not a subset of the union" or "is not equal to the union" or perhaps "0 and 1 are not members of the union". –  Trevor Wilson Feb 8 '13 at 22:30

3 Answers 3

If $I$ is some index set and for $i\in I$ we have a set $A_i$ then $x\in\bigcup_{i\in I} A_i$ if and only if there exists some $i\in I$ such that $x\in A_i$.

The difference between $[0,1]$ and $(0,1)$ is the endpoints, namely $0,1$ are in the closed interval but not in the open. $0$ would be in the union if and only if it is an element in one of the set we take union over, namely $\left[0+\frac1{n+1},1-\frac1{n+1}\right]$. But neither $0$ nor $1$ are elements of any such interval, so they cannot be in the union.

This shows that the increasing union of closed intervals need not be a closed interval. But it is also generally the case that the union of infinitely many closed sets is not closed.

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Because $0$ and $1$ are not members of any of the intervals. Taking the limit is like taking the closure of the set, and you are correct that the closure of $(0,1)$ is $[0,1]$. But for any number in $(0,1)$ we can find a particular $n$ that is one of the sets it is in (it is also for any greater $n$). Not for $0$ or $1$.

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$[0,1]$ is not a member of the union, because there does not exist any $n$ such that

$$ 0 = 0 + \frac{1}{n+1} \qquad \text{and} \qquad 1 = 1 - \frac{1}{n+1} $$

It's not contained in the union because $0 \in [0,1]$, but there does not exist an $n$ such that

$$ 0 + \frac{1}{n+1} \leq 0 \leq 1 - \frac{1}{n+1} $$

(and similarly for $1$)

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Technically the first sentence shows that $[0,1]$ is not a member of the set whose union is being considered. ($[0,1]$ is also not a member of the union itself, simply because it is not a real number.) –  Trevor Wilson Feb 9 '13 at 0:08

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