Cute Proofs: Closure of a Connected Subspace is Connected.

May 16, 2011

[Big Edit: Thank you to the commenters below (and especially the latest one) who pointed out some of my previous statements only held true in a specific case.  I’ve streamlined the proof, and I think everything is kosher now.  I’ve also added some counterexamples to the end.]

I was reading through some of the earlier chapters of Bredon and I got to one of those questions that should be easy but I couldn’t recall the argument off of the top of my head.  The actual question is slightly more general, but the question we’ll answer first is:

Problem: If A is connected, show \bar{A} is connected.

 

There are several nice definitions of connected to choose from.  In particular, we have:

A subset A\subseteq X is connected if one (and hence all) of the following hold:

  • There DO NOT exist disjoint non-empty closed (with respect to A) sets C,D such that C\cap A\neq\emptyset and D\cap A\neq \emptyset but C\cup D = A.
  • The only subsets of A which are both open and closed in the relative topology are \emptyset and A.

 

[Note:  In a former draft of this post, I had (incorrectly!) stated that a subset A is connected if we did not have two disjoint open sets which do not intersect A trivially, but which union to A.  This can be shown to be false (which was done by a clever commenter below), but it will work if X (and therefore A in the subspace topology) is a metric space.]

 

To solve this, we will use the first statement above but, for fun, try to prove it yourself using the second statement.  We’ll be using the contrapositive, meaning we will prove if \bar{A} is not connected, then A is not connected.  This is equivalent to proving the theorem above (why?).

 

Solution.  Let it be the case that \bar{A} is not connected, so that there exists C,D disjoint closed sets neither of which intersect \bar{A} trivially.   Suppose that A is connected; then for any two disjoint closed sets C,D with A\subseteq C\cup D we have that one of these closed sets intersects A trivially (or else it would not be connected).  In other words, wlog we have A\subseteq C.  Taking the closure of both sides, we have \bar{A}\subseteq \bar{C} = C, but since C and D are disjoint this implies that \bar{A}\cap D = \emptyset, contradicting the fact that neither C nor D may intersect \bar{A} nontrivially.  \Box

 

In fact, nearly the same kind of argument shows that if there is some B such that A\subseteq B\subseteq \bar{A} with A connected, then B is also connected.  This is pretty nice; it tells us we can add limit points to our connected space (even if we only add some limit points and not others!) and it remains connected.

 

Some counter examples…

One might ask, "Hey, what if \bar{A} is connected?  Then is A connected?"  And a minute of thinking should provide a counter-example.  If not, then just think about A = {\mathbb Q} and its closure.

Additionally, one might think that if a set A is connected, then its interior (if it is nonempty)  would be connected.  This is true in the case of the closed disk D\subset {\mathbb R}^{2}, but if we were to put two disjoint disks in the plane and connect them together with a line (this will look like a barbell, or a pair of eyeglasses) the interiors will be two disjoint open disks, which are clearly not connected. 

Maybe, though, the boundary of a connected set A is always connected (if it is nonempty).  To do away with this idea, just think about the interval in {\mathbb R}.

A question to think about (assume all of these sets are non-empty):

  • If the boundary of some set A is connected, is A connected?  (For this one, think about a figure-8 in the plane.)
  • If the interior of some set A is connected, is A connected?  (For this one, think about what the interior of a point is, and then let A be some nice set with non-empty interior union some point.)
  • If the interior and boundary of some set A are connected, is A?  (For this one, you might have to think a bit.  I’ll try to hide an answer a little bit so you can work on it.  Consider the set of positive rational numbers, {\mathbb Q}_{+} union the interval [0,1].  The interior will be (0,1) which is connected, and the boundary will be [0,\infty) which is connected,  but the set itself is not connected (for the same reason the rationals are not connected). 
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11 Responses to “Cute Proofs: Closure of a Connected Subspace is Connected.”

  1. The argument I came up with that I always liked uses the fact that a space $atex X$ is connected if and only if there does not exist an epimorphism (continuous surjection) from X to the two point discrete space \bold{2}.

    Now, let A\subseteq X be connected and let f:X\to \bold{2} be continuous. Now, since f is continuous and A connected we know that f_{\mid A} is constant, and since the extended constant map g:X\to\bold{2} (sends everything in x to the constant things in A are sent to) is continuous. But, recall that if k:E\to H is a continuous map where E is dense in Y and H Hausdorff then there is at most one extension \widetilde{k}:X\to H. So, in particular for our problem we must have that g=f. Thus, every continuous map f:X\to \bold{2} is constant and so X is connected.

    Really the proof is much shorter, I just put in all the details.

  2. Haha, that got a little garbled. It should really read

    Now, let A\subseteq X be connected and let f:X\to \bold{2} be continuous. Now, since f is continuous and A connected we know that f_{\mid A} is constant, and the extended constant map g:X\to\bold{2} (sends everything in x to the constant things in A are sent to) is continuous. But, recall that if k:E\to H is a continuous map where E is dense in Y and H Hausdorff then there is at most one extension \widetilde{k}:Y\to H. So, in particular for our problem we must have that g=f. Thus, every continuous map f:X\to \bold{2} is constant and so X is connected.”

    • James said

      I think Bredon introduces connectivity like this and this proof is pretty slick, no lie. Most of the work, I feel, is in the “extension” part which is a nice little fact in its own right.

      Also, I should comment that I began reading your blog and I enjoy it! Nice work!

      • Thanks man, I like your blog as well! I’ve done mostly representation theory of finite groups, group theory, and some point-set topology on mine–I hope to do ring theory/commutative algebra and multivariable analysis/differential topology/differential geometry soon…hopefully you find the latter enjoyable

  3. Also, I find this kind of ironic, but another proof that I was always proud of is that statement you said ‘contains the most work’. Namely, that if D is dense in X and k:D\to H is continuous where H is Hausdorff then there is at most one extension \widetilde{k}:X\to H. Indeed, it will clearly follow if we can prove that for any two continuous f,h:X\to H the set A(f,g)=\left\{x\in X:f(x)=g(x)\right\} is closed (since it contains the dense set D. But, the fact that A(f,g) is closed follows by recalling that since H is Hausdorff the diagonal \Delta_H\subseteq H\times H is closed–and noticing that f\oplus g:X\to H\times H:x\mapsto (f(x),g(x)) is continuous (since each coordinate map is continuous) and finally noticing that A(f,g)=\left(f\oplus g\right)^{-1}\left(\Delta_H\right)

  4. Brooke Ullery said

    jmy, in your definition, you say U and V are closed, but then in your proof you treat them as though they are open. Which do you mean? If they are closed, then A \subset U obvs implies A closure \subset U.

    • James said

      Thanks, brk. I’ve fixed some typos and I modified the proof a bit. I don’t even know why I needed that last property, since the proof really is just the first one. Worst.

  5. beroal said

    “A subset $A\subseteq X$ is connected if one (and hence all) of the following hold: There DO NOT exist disjoint non-empty open sets $U,V$ such that $U\cup V = A$.”
    You forget to mention that $U,V$ are open in $A$. If you want to consider openness in $X$, then the definition will be “There DO NOT exist open sets $U,V$ such that $U\cup V = A$ and $U\cap V\cap A = \varempty$ and $U$ meets $A$ and $V$ meets $A$”.

    “Solution. Suppose that $A$ is connected. Now suppose that $\bar{A}\subset U\cup V$ for $U,V$ are disjoint open sets. … V^{C} is the compliment of V. But this is a closed set!”
    I believe that the hypothesis should be (based on your definition): “Now suppose that $\bar{A} = U\cup V$ for $U,V$ are disjoint, open in $\bar{A}$ sets”. Then “$V^{C}$ is closed” (I assume, closed in $X$) is not justified. Something is wrong.

    • James said

      In this case, I actually wanted them to be open in X, which is crucial but I did not point out. The additional criteria are also necessary, as you pointed out; I’ve edited it to include those. Now that I have this additional stipulation on U,V being open in X, we would have V^{C} closed in X (and therefore in the subspace topology on A). Thanks for pointing this out!

      • beroal said

        Then your definition differs from the standard definition. Consider a set {0, 1, 2} with a topology {{}, {0}, {0, 1}, {0, 2}, {0, 1, 2}}. {1, 2} is disconnected w.r.t. the standard definition (consider {{0, 1}, {0, 2}}) and is connected w.r.t. your definition (A=\{1, 2\}, U\cup V = A, U\subseteq A, V\subseteq A, every open set included in \{1, 2\} is empty, so U, V are both empty).

      • James said

        You’re right — it looks like my definition holds if the space is a metric space and possibly for some weaker spaces, but [as your example shows] not for all. That’s a bit upsetting. I will update this post soon to reflect this, but I need to figure out if I want to add the assumption that X is a metric space or if I want to try something else.

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