# Reference for showing two closed and convex sets are equal

Suppose I have two sets $P, Q \subseteq R^d$ such that $P\subseteq Q$. $P$ and $Q$ are both convex and closed. I wish to show that $P=Q$.

A straightfoward way to show this is showing $\forall y \in R^d$, $y \notin P \Rightarrow y \notin Q$. However, I follow a slightly different technique which utilizes the closedness and convexity of both the sets which I believe is known as the continuity method in topology (although I may be wrong.)

I can show in my problem that for any $x \in P$, $\exists \epsilon$ such that $\forall y \in B(x,\epsilon)$ (closed ball of radius $\epsilon$ around a point $x \in R^d$) I have, $y \notin P \Rightarrow y \notin Q$ i.e. all points immediately outside the boundary of $P$ don't belong to $Q$ as well.

I believe this is sufficient to show $P=Q$. Is this fine? If yes, does anyone have a reference I can cite for this (this is probably not very deep but the publication is applied and hence a reference is needed.) Reference at the level of appropriate theorems would be most appreciated.

• There must be some extra hypothesis missing as the closed ball of radius 1 is not equal to the closed ball of radius 2. – azarel Feb 9 '12 at 4:50
• It looks to me like he's arguing that $Q\subset\overline{P}\Rightarrow P=Q.$ The hypotheses don't hold for your example. – Neal Feb 9 '12 at 4:53

It seems sufficient if $P$ is nonempty and closed and $Q$ is convex.

Suppose there is $a \in Q \setminus P$, and let $b \in P$. Let $m = \max \{ t \in [ 0, 1 ] \mid b + t(a - b) \in P \}$. This maximum exists and is less than 1 because $P$ is closed.

Now $b + m(a-b) \in P$ and for every $\epsilon \in (0, 1-m)$ we have by convexity $b + (m+\epsilon)(a - b) \in Q \setminus P$.

Of course this can easily be generalized from convex to path-connected, and merely connected should still work. If you want a reference, you will probably only find more general results, so you might as well try to fit the above in a footnote, if necessary.

• I think the topological version is cleaner. Essentially if $P$ and $Q$ are subsets of $X$ such that $Q$ is connected, one has that if there exists and open set $\Omega$ such that $P\subseteq\Omega\subseteq\bar{\Omega} \subsetneq X$ then $P = Q\cap \Omega \implies P = Q$. – Willie Wong Feb 9 '12 at 19:50
• @Willie Wong: I agree that the topological argument is more elegant, but it's also more abstract. I don't know which is more appropriate for the target audience. – Niels J. Diepeveen Feb 9 '12 at 20:56

The method of continuity probably should be stated slightly differently (not because you stated it wrong; just that a slightly different statement may make it more evident that you are using the method of continuity, so you don't have to find a reference book).

Continuity method says that if $Q$ is a connected topological space, and $P\subseteq Q$ is nonempty. If $P$ is both open and closed in $Q$ then $P= Q$.

That $Q$ is connected follows from convexity. $P$ is nonempty by assumption. That $P$ is closed follows from the fact that you are using the topology on $Q$ that is induced by the topology on $\mathbb{R}^d$. So you just need to check that $P$ is open. Using the induced topology, we can state it as: it suffices the check that for all $p\in P$ there is an open neighborhood $N_p$ in $\mathbb{R}^d$ of $p$ such that $N_p\cap Q \subseteq P$.

What you've been able to show (in your question statement) is precisely that for every open ball $B(p,\epsilon)$ around a point $p\in P$, you have that $B(p,\epsilon) \setminus P \subseteq B(p,\epsilon)\setminus Q$. Which by taking the relative complement is precisely $B(p,\epsilon)\cap Q \subseteq B(p,\epsilon)\cap P\subseteq P$ as you desired. So yes, deep down, what you did is the method of continuity.

This is a more-or-less straightforward point-set topology problem in $\mathbb{R}^d$, so you probably won't find a literature reference. You might try looking in Munkres' Topology, or the point-set introduction to an analysis book such as Rudin's Principles of Mathematical Analysis.

To the heart of the matter. Your claim does not even require convexity. Rephrased,

Suppose $Q$ is a closed subset of $\mathbb{R}^d$ and $P$ is a closed subset of $Q$. Suppose further that for all $\epsilon>0$ and any $x\in P$, if $y\notin B(x,\epsilon)$ then $y\notin Q$. Then we may conclude that $P=Q$.

Proof: It suffices to show that $Q\subseteq P$. Let $y\notin P$; we show that $y\notin Q$. Because $y\notin P$ and $P$ is closed, hence $P'$ is open, there exists some positive $\eta$ such that $\inf_{x\in P}d(x,y)>\eta$. But now by hypothesis we have $y\notin Q$. Therefore $Q\subseteq P$ (by contrapositive); since also $P\subseteq Q$, we may conclude that $Q=P$.

• Thanks azarel and neal. @Neal: thanks for nifty proof. However, I need a weaker condition than what you have stated to show the same thing. I wish to show $\mathbf{\exists \epsilon}$ such that for any $x \in P$, if $y\notin B(x,\epsilon)$ then $y\notin Q$. That is in effect, I care only about the boundary of $P$ as for any point $x$ in the relative interior of $P$, it is trivial to pick a small enough $\epsilon$ such that $y \notin B(x,\epsilon)$ is not true (hence the claim is trivially true.) I suppose this is where convexity is needed. – Rajhans Feb 9 '12 at 16:34
• @Azarel: $B(0,1) != B(0,2)$ in my claim because for a point $x$ on the boundary of $B(0,1)$, there is no such $\epsilon >0$ such that for all $y$ in an epsilon ball around $x$, $y \notin B(0,1) \Rightarrow y\notin B(0,2)$ as any point immediately outside the boundary of $B(0,1)$ lies in $B(0,2)$. – Rajhans Feb 9 '12 at 16:34