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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 $n$? –  MJD Oct 18 '12 at 3:29

2 Answers 2

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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Yep, thank you. –  DonAntonio Oct 18 '12 at 3:31
    
Can't understand why someone would downvote an obvious typo. At least I can cancel it! –  André Nicolas Oct 18 '12 at 3:33
    
Thanks @André. I don't get very annoyed by this any more. Some people just seem to enjoy downvoting, so let knock themselves out. –  DonAntonio Oct 18 '12 at 3:37

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