Infinite Intersections Of Open Sets

Geometry Level 3

Let $U_{\alpha}$ $(\alpha \in A)$ be a collection of open sets in ${\mathbb R}^2.$ If $A$ is finite, then the intersection $U = \bigcap\limits_\alpha U_{\alpha}$ is also an open set. Here is a proof:

Suppose $x\in U.$ For each $\alpha \in A,$ let $B_{\alpha}$ be a ball of some positive radius around $x$ which is contained entirely inside $U_{\alpha}.$ Then the intersection of the $B_{\alpha}$ is a ball $B$ around $x$ which is contained entirely inside the intersection, so the intersection is open.

$($ Here a ball around $x$ is a set $B(x,r)$ ( $r$ a positive real number) consisting of all points $y$ such that $|x-y|<r.$ In ${\mathbb R}^2$ it is an open disk centered at $x$ of radius $r.)$

Where does this proof go wrong when $A$ is infinite ?

The intersection of infinitely many sets is not necessarily defined The proof works; the statement is true

B

might not be contained inside

U

U

might be empty

B

might not be a ball around

x

1 solution

Patrick Corn
Jun 7, 2016

Relevant wiki: Open Sets

When $A$ is finite, the intersection of the balls $B(x,r_\alpha)$ is $B(x,r),$ where $r$ is the minimum of the $r_\alpha.$

When $A$ is infinite, there is not necessarily a minimum of the $r_\alpha.$ In fact, the intersection $B$ might very well be the one-point set $\{x\},$ which is not a ball around $x.$ For instance, the intersection of open sets $B(x,1) \cap B(x,1/2) \cap B(x,1/3) \cap \cdots$ equals the one-point set $\{x\},$ which is not open. So the correct answer is that $B$ might not be a ball around $x.$

(Note that the other choices are not correct: the result is certainly false, as seen above; $B$ is indeed contained in $U$ ; $U$ might be empty, but that's not a problem since the empty set is (vacuously) open; and the intersection of infinitely many sets is certainly defined.)

$U$ being empty is a problem with the proof, even in the finite case. Even in the finite case, the proof should go:

"If $U$ is empty, then $U$ is open. If $U$ is nonempty, then for any $x \in U$ ..."

As it stands, the proof does not deal with the $U = \varnothing$ case at all.

Mark Hennings - 5 years ago

The condition that a set $U$ is open says that "for any $x\in U,$ some condition is satisfied." If $U$ is empty, the statement in quotes is vacuously true since there is no $x\in U.$

In other words, the proof shows that for any $x\in U,$ there is a ball around $x$ contained in $U.$ This is true whether or not $U$ is empty. It is not necessary to set aside $U=\varnothing$ as a special case.

Patrick Corn - 5 years ago

Doesn't this question belong to the calculus section?

A Former Brilliant Member - 4 years, 12 months ago

Not sure. I will ask.

Patrick Corn - 4 years, 12 months ago

In the absence of a Topology section, it is a close call. Calculus is probably more correct, but there are separate strands in Functional Analysis studying (a) functions on metric spaces, and (b) their geometry.

Mark Hennings - 4 years, 12 months ago

Infinite Intersections Of Open Sets

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