Finite intersection of open sets is open

  • Thread starter cappygal
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In my multivariable calc class, we're asked to prove that the finite intersection of open sets is open. I've tried to find help on the internet but couldn't find anything to help. I understand somewhat the idea of "nesting sets" that some proofs use .. can anyone help me understand this to prove it? Thanks!


Science Advisor
It would help if you could provide some context. What is the space you are working in and how are open sets defined? In purely abstract situations, the set of open sets is closed under finite intersections and all unions by definition.

It is a purely abstract situation. The book says: "Prove the following statements for open subsets of R^n (where R is the Reals):
a) any union of open sets is open.
b) A finite intersection of open sets is open
c) An infinite intersection of open sets is not necessarily open. "

I can prove a and c, but not b. My book is Vector Calculus, Linear Algebra, and Differential Forms, by John H and Barbara Burke Hubbard. Since it is not a topography book, it does not explain any of this within the context of the chapter reading. I hope this helps clarify.


Science Advisor
Homework Helper
That's not a purely abstract situation. That's a very specific situation, where your space is Rn. Normally, open sets of some set X are defined to be the elements of a topology T, where T is a collection of subsets of X satisfying the following properties:

1) the empty set and X itself are both elements of T
2) every arbitrary union of elements of T is itself an element of T
3) every finite intersection of elements of T is itself an element of T

So topologically speaking, by definition, a finite intersection of open sets is open, since "being open" just means "being an element of the topology." Note that a set X (where X might be some Rn, or possibly anything else) can have various different topogies. You are studying the standard topology, I assume.

You are probably defining an open set to be a set U such that for each x in U, there is an open ball B such that x is in B and B is a subset of U. This is an even more specific case (less abstract) since you're working within a specific topology. Moreover, you're probably appealing to a metric (open balls around x are the set of points that are within some given distance from x). In general, you don't need a metric to define a topology. Even more specific is that you're dealing with a particular metric, the Euclidean metric. So open balls around x will really "look like" balls. With other metrics, they can look like other things, like cubes perhaps. In fact, you might be dealing with open rectangles instead of open balls.

Well you can prove this by induction. If you can prove that the intersection of two open sets is open, then you can prove that a finite intersection is open. If you're working with open balls, you want to prove that if x is in two balls U and V, then there is some open ball centered at x which is contained in both U and V. So if you think about drawing two overlapping circles, then at any point where they overlap, you can draw a tiny circle around that point so that this tiny circle is entirely contained in the overlapping region. If you're dealing with rectangles, its even easier because the overlapping region is itself a rectangle, so its very easy to choose the rectangle (just choose the intersection!). Now this should give you an intuitive feel for what you should be trying to prove. But you will have to keep in mind that you're dealing with Rn[/sub], not just R², so you'll have to generalize these ideas. You'll have to show rigourously that the intersection of two open rectangles is again an open rectangle. If you're dealing with open balls, you have to explicitly choose a radius r for your tiny ball to ensure that it is contained in the intersection of U and V. These things are easy to do once you have a clear picture of what you're trying to do.

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