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If I show this statement:

$$x\in \left] a,b \right[ \Rightarrow \exists n \in \mathbb{N} : x\in \left] -\frac1{n}, 1+\frac1{n}\right[$$

Have I then shown this statement:

$$]a,b[ \subseteq \bigcup_{n=1}^\infty \left] -\frac1{n}, 1+\frac1{n}\right[\qquad ?$$

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Yes. And conversely if you have proved the second, you have proved the first. Whether the two statements are "the same" is a more complicated matter. It all depends on the exact definitions that have been given for $\subseteq$, for $\cup_{n=1}^\infty$, and so on. The two statements are likely not the same, though they certainly are equivalent. – André Nicolas Oct 4 '11 at 15:45
What do the opposite brackets mean? – mixedmath Oct 5 '11 at 1:58
up vote 1 down vote accepted

Note that:

  • A set $A$ is a subset of a set $B$ if and only if $x\in A\Rightarrow x\in B$.

  • If we have $A_i$ then $x\in\bigcup A_i$ if and only if for some $i$ we have $x\in A_i$.

Combine the two result and you have indeed what you wanted.

Note, while at it, that $(-1,2)$ which is the interval for $n=1$ (since $-\frac{1}{1}=-1,\ 1+\frac{1}{1}=2$) is a superset of all the other intervals. In particular this whole union is just $(-1,2)$ to begin with.

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Actually, as the supremum of each interval is $1+1/n$, not $1/n$, the entire union is $(-1, 2)$, not $(-1, 1)$. – jwodder Oct 4 '11 at 16:59
@jwodder: Thanks for the correction. – Asaf Karagila Oct 4 '11 at 17:39

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