# Why is infinite intersection “towards infinity” an empty set?

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.

• The notation $\bigcap_{i=1}^\infty F_n$ is wrong... – David C. Ullrich Sep 8 at 15:25

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

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.

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 $$F = \emptyset$$.

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.

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.