Nonempty convex sets

1. Dec 31, 2005

Hi,

Let $$C_1$$ and $$C_2$$ be nonempty convex sets and suppose $$C_1 \cap C_2 \neq \emptyset$$. I read a text that claims $$\textup{cl}(C_1 \cap C_2) \subset \textup{cl}(C_1) \cap \textup{cl}(C_2)$$ since $$C_1 \cap C_2 \subset \textup{cl}(C_1) \cap \textup{cl}(C_2)$$.

I am able to prove the latter, but I am not able to see the deduction made by the author. Hope someone can help me out.

Let $$x \in C_1 \cap C_2 \neq \emptyset$$, then $$x \in C_1$$ and $$x \in C_2$$. This implies $$x \in \textup{cl}(C_1)$$ and $$x \in \textup{cl}(C_2)$$. Hence $$x \in \textup{cl}(C_1) \cap \textup{cl}(C_2)$$. So $$C_1 \cap C_2 \subset \textup{cl}(C_1) \cap \textup{cl}(C_2)$$.

p/s: cl means closure

Last edited: Dec 31, 2005
2. Dec 31, 2005

HallsofIvy

Staff Emeritus
Don't worry about x being a member of $C_1\capC_2$. Let x be a limit point of $C_1\capC_2$. Then there exist an open neighborhood U of x which contains at least one point of $C_1\capC_2$. Therefore, that same neighborhood contains at least one $C_1$ and so x is a limit point of C_1 and so x is contained in $Cl(C_1)$. Now do the same thing for C2.

3. Dec 31, 2005

mathman

I am not sure how much you covered in the theory of closed sets. However, one very early theorem is that all intersections of closed sets (as well as all finite unions) are closed. Therefore the intersection (K) of the closures of C1 and C2 is closed, so that if C1 int. C2 is inside K, then its closure must be inside K.

4. Dec 31, 2005

benorin

The closure of a set is the union of the set itself and the set of its limit points (e.g. the dervired set), say $\mbox{cl}(C)=C\cup C^{\prime}$; hence

$$\textup{cl}(C_1) \cap \textup{cl}(C_2) = \left( C_1 \cup C_1^{\prime}\right) \cap \left( C_2 \cup C_2^{\prime}\right) = \left( C_1 \cap C_2\right) \cup \left( C_1^{\prime} \cap C_2\right) \cup \left( C_1 \cap C_2^{\prime}\right) \cup \left( C_1^{\prime} \cap C_2^{\prime}\right)$$

(this follows upon two applications of the identity $A \cap \left( B \cup C\right) = \left(A \cap B \right) \cup \left(A \cap C \right)$,)

therefore

$$C_1 \cap C_2 \subset \textup{cl}(C_1) \cap \textup{cl}(C_2)$$

Last edited: Dec 31, 2005
5. Jan 1, 2006

matt grime

I think I prefer the categorical idea that the closure of U is 'universal with the property' ie it is the smallest closed set containing U, and is thus contained in any other closed set that contains U.

6. Jan 1, 2006

I am looking for a proof of

$$C_1 \cap C_2 \subset \textup{cl}(C_1) \cap \textup{cl}(C_2) \Longrightarrow \textup{cl}(C_1 \cap C_2) \subset \textup{cl}(C_1) \cap \textup{cl}(C_2)$$

and not

$$C_1 \cap C_2 \subset \textup{cl}(C_1) \cap \textup{cl}(C_2)$$

Last edited: Jan 1, 2006
7. Jan 1, 2006

matt grime

We know, and there are what, 3 different proofs of differing complexity (but all the same, relying on the characterization of the closure of a set as the intersection of all closed sets containing it, which I believe gives you a 4th proof now). The simplest is probably: A in B implies cl(A) is in cl(B), and if C is closed the cl(C)=C, which is mathman's.

Last edited: Jan 1, 2006
8. Jan 5, 2006

I am able to prove intersection of closed sets is closed. I see that you are just claiming if $$C_1 \cap C_2 \subset K$$ then $$\textup{cl}(C_1 \cap C_2) \subset K$$, without proving it.

Last edited: Jan 5, 2006
9. Jan 5, 2006

mathman

I guess I should have mentioned a basic theorem. The closure of a set A is the smallest closed set containing A. From this it is obvious that any closed set containg A will also contain the closure of A.

10. Jan 5, 2006

HallsofIvy: I find your method quite elegant.

Benorin: Thank you for an alternative way for showing $$C_1 \cap C_2 \subseteq \textup{cl}(C_1) \cap \textup{cl}(C_2)$$.

Ah! I see it now (after going over it a couple of times). Thank you.

Just after I've posted my last reply, I came up with an idea, although rather long. (it gives me better insight to the problem)

Given two sets C and D. I shall prove $$\textup{cl}(C) \subseteq \textup{cl}(D)$$ provided that $$C \subseteq \textup{cl}(D)$$.
Suppose this is not the case, i.e. $$\textup{cl}(C) \nsubseteq \textup{cl}(D)$$, then there must exist $$x \in \textup{cl}(C) \backslash \textup{cl}(D)$$. Clearly this x does not belong to $$\textup{cl}(D)$$. It follows that there exists an open ball of x with radius $$\epsilon > 0$$ sufficiently small such that the ball does not cover any points belonging to $$\textup{cl}(D)$$. Such ball exists since for otherwise x would have belonged to $$\textup{cl}(D)$$.$$^1$$ The ball contains some points belonging to C that reside in the exterior of $$\textup{cl}(D)$$. But this contradicts with the assumption that $$C \subseteq \textup{cl}(D)$$.

I welcome some feedback.

$$^1$$ However, the existence of an open ball may not be true had we generalized $$\textup{cl}(D)$$ to $$D$$ even if $$x \notin D$$.

Last edited: Jan 5, 2006
11. Jan 6, 2006

matt grime

Why are you doing this the hard way? How about you define for me the closure of a set? For me it is the intersection of all closed sets containing it. This makes every question you've asked absolutely trivial. Therefore you must be adopting a definition different from mine. But then you appear to be assuming that balls of radius e make sense so you can only be operating in a metric space. I doubt anyone else here was making that assumption to prove the result you need which is true in any topological space that if U is in K and K is closed then the closure of U is in K too.

http://mathworld.wolfram.com/SetClosure.html

are (some of) the equivalent definitions there are.