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So, the representation of this is simple:

${\{x \in \mathbb R | -1 \le x \le 1/n\}}$

But I'm not sure what I need to prove. This is all of the information that I have. In class, a different problem, we were told what the set was equal to, and had to prove/disprove that. So, should I prove $\cup_{n = 1}^\infty [-1, 1/n] = [-1, 1]$?

Or is there something else that I need to do?

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    In your simple representation ${\{x \in \mathbb R | -1 \le x \le 1/n\}}$, what is the value of $n$?2012-10-18

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You’ll have a hard time proving that $\bigcup_{n = 1}^\infty \left[-1, \frac1n\right] = \left[-1, \frac1n\right]\;:$ it doesn’t even make sense. The $n$ on the lefthand side is a dummy variable: the value of the expression wouldn't change if you replaced it by something else, say $k$, to get $\bigcup_{k = 1}^\infty \left[-1, \frac1k\right]\;.$ The $n$ on the righthand side, however, is apparently a particular integer. Thus, you’re using one letter, $n$, to represent two unrelated things of very different kinds.

You’re actually being asked to figure out exactly what real numbers are in the set $\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]$ and then to express the set in a way that doesn’t require talking about infinitely many sets. For example, it should be clear that every real number in the interval $[-1,0]$ belongs to the set. However, $\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]\ne[-1,0]\;,$ because, for instance, $\frac13\in\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]\;,$ but $\frac13\notin[-1,0]$.

Sketch the intervals $\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]$ for the first few positive integers $n$ to get an idea of what they look like. Once you’ve done that, figure out exactly what the set $\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]$ looks like, and write down a simple description of that set. Finally, if your set is $A$, prove that $\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]=A\;,$ probably by showing that if $x\in A$, then $x\in\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]$, and if$x\in\bigcup_{n = 1}^\infty \left[-1, \frac1n\right]$, then $x\in A$.

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I think you're asked to represent the infinite union in a closed way:

$\bigcup_{n=1}^\infty \left[-1\,,\,\frac{1}{n}\right]=[-1\,,\,1]$

Can you see why?

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    Thanks @André. I don't get very annoyed by this any more. Some people just seem to enjoy downvoting, so let knock themselves out.2012-10-18