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Why is the infinite intersection "towards infinity" an empty set?

Or i.e.

Why is:

$$\cap_{i=1}^{\infty} F_i = \emptyset$$

$$F_n=[n, \infty)$$

There's intuition, the intersection is always the "smallest of the sets" so eventually it will be $(\infty,\infty)$ or something like that.

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  • $\begingroup$ The notation $\bigcap_{i=1}^\infty F_n$ is wrong... $\endgroup$ Sep 8, 2019 at 15:25

5 Answers 5

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To belong in the intersection, any element would have to belong in each of the sets $[n,\infty)$ which means it must be larger than every finite number. Since there is no finite number with this property, the intersection is therefore empty.

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The intersection is made up of real numbers which are greater than or equal to every positive integer. By Archimedes' property, there's none.

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Suppose, to obtain a contradiction, that $F=\cap_n F_n$ is non-empty. Let $x \in F$. Then, $x\ge n$ for all $n \in \mathbb N$. However, there is no largest real number, so we must conclude that $F = \emptyset$.

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Let's follow your intuition:

There's intuition, the intersection is always the "smallest of the sets" so eventually it will be $(\infty,\infty)$ or something like that.

Okay, but $(\infty,\infty)$ doesn't have any elements in it. Assuming we're dealing with the domain of real numbers, $\infty$ isn't one. (Recall that in real numbers the symbol $\infty$ isn't a number; it's always a shorthand placeholder for some statement about unbounded limits.)

On the other hand, even if we were talking about the extended real numbers where $\infty$ is an element, the notation $(\infty,\infty)$ indicates that endpoints are not included; that being $\infty$, so again this is an "interval" with no points in it.

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It might be a little easier to understand via the contrapositive:

Let $x$ be any real number. Then we know that there exists a positive integer $n_x$ larger than $x$. (This is the so-called Archimedean property of the reals, but intuitively you can think of $n_x$ just being $x$ rounded up to the next integer, or if $x$ is negative you can just let $n_x$ be $1$.) That means $x$ is not in the set $F_{n_x}$, so it certainly cannot be in the intersection $\bigcap_n F_n$. This is true no matter what $x$ is, so $\bigcap_n F_n$ cannot contain any real numbers at all, which is to say it equals the empty set.

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