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This is probably a really stupid question, but it's been annoying me for a while and I still can't find an answer that convinces me.

We know that $\lim_{x \rightarrow \infty} \frac{1}{x} = 0$.

But, in my lecture, we saw that $\{0\} = \bigcap_{n=1}^{\infty} \left(-\frac{1}{n} , \frac{1}{n} \right) \subset \mathbb{R}$.

What I don't get is, why isn't the above intersection equal to $(0,0)$, i.e. to the empty set? Again, sorry if this is really stupid. Someone told me it's because $\frac{1}{n}$ approaches 0 but is never equal to $0$, but then why would $\lim_{x \rightarrow \infty} \frac{1}{n}$ be equal to 0?

Thanks in advance for any help!

justdoit
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3 Answers3

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The intersection of an infinite amount of sets is defined to be the elements present in every set.

$0$ is in every set of the form $\left( -\dfrac1n, \dfrac1n \right)$, so it is also in their intersection.

There is no rules saying that arbitrary intersections of open set must be open, and this is a counter-example.

Kenny Lau
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Do not apply limit to the numbers, what $ \bigcap_{n=1}^{\infty} \left(-\frac{1}{n} , \frac{1}{n} \right) \subset \mathbb{R}$ means is that it is a set of all element that is contained by all sets of $(-1/n, 1/n) $ for all $n \in \mathbb N$.

Thus only $0$ satisfies it.

Notice we could naturally define arbitrarily number of set intersections and unions, even uncountably many is fine, thus no limit is needed - the $\infty$ in your formula is just a notation saying that we take $n$ for all natural numbers.

Jay Zha
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  • Thanks for your answer ; however it's still a little bit fuzzy in my head (because we could still take $n$ extremely big), so here is another question: could it be that $\bigcap_{n=1}^{\infty} = {0}$ comes from the fact that, since $\mathbb{N}$ is an infinite set, even if we chose a really big number $n_0$, we could still find another bigger number $n_1 > n_0$, and therefore we never really reach $\infty$? – justdoit May 10 '17 at 10:30
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    @alissad Yes, you are right, the point here is that $n$ cannot reach $\infty$ - so $\infty \notin \mathbb N$, and $n = 0 $ to $\infty$ here is just a notation for $n \in \mathbb N$, so $n$ could not take $\infty$ – Jay Zha May 10 '17 at 11:36
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Because the limit is the value where there are an infinity of intervals "around". In this case you can choose any $\epsilon>0$ in which $1/x_{n}$ stays in $(0-\epsilon,0+\epsilon)$ for all $n> N(\epsilon)$ (when $x_{n}$ goes to $\infty$).

And so, the intersection always contain $0$ because it is an open interval (neighbourhood of $0$).

thewatcher
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