# Gracious Living

Connectedness
November 9, 2010, 09:30
Filed under: Math, Topology | Tags: , ,

Taking a break from the construction of topologies to talk about some of their major properties.  First on the docket is connectedness.  This is something that everybody’s got an intuitive grasp on, though the intuition does have some interesting deviations from the norm.

A connected space is one that cannot be written as a union of open sets $A$ and $B$ with $A\cap B=\emptyset$.  Note that $A$ and $B$ in this definition would be complements of each other, hence closed as well; on the other hand, given a clopen set(=”closed”+”open”, remember), its complement is also clopen, so the space containing them is disconnected.  So an equivalent definition is: a space is connected iff it has no clopen sets.  Also, the fact that the definition is stated in terms of a pair of sets is irrelevant — given any collection of pairwise disjoint open sets whose union is the whole space, we can just take $A$ to be a single one of those sets, and $B$ to be the union of the rest of them, which is still open.  We’ll call such a collection of sets a separation of the space.

This also gives us an equivalence relation on a topological space: two points are equivalent if there is a connected subspace containing them.  The equivalence classes are then the “maximal” connected subspaces: we call them connected components.

Some examples are in order.  Any set with the discrete topology (and, um, with more than two points) is disconnected — in fact, it is totally disconnected, since every point is in a different connected component.  But total disconnectedness doesn’t imply a discrete topology: for example, the subspace topology on $\mathbb{Q}$ is not discrete, but given two points, we can find a separation by just choosing an irrational number $\alpha$ between them and using the rays $(-\infty,\alpha)\cap\mathbb{Q},(\alpha,\infty)\cap\mathbb{Q}$.  Similarly, the lower limit topology on $\mathbb{R}$ is not discrete, but given $a, the rays $(-\infty,b),[b,\infty)$ are a separation.

From the planet of Let’s Do Something Goofy, the empty set is disconnected in its only topology: just let $A=\emptyset, B=\emptyset$ and you’ve got your separation.

With order topologies, one thing to do is look for separations into two rays, which means that if some closed ray $[a,\infty)$ or $(-\infty,a]$ is open, the space is disconnected.  This happens, for example, if $a$ has a successor, that is, a smallest element greater than it.  Then $[a,\infty)=(S(a),\infty)$, which is open.  On the other hand, the space $(0,1)\cup\{2\}$ with the order topology is connected — in fact, it’s homeomorphic to $(0,1]$!  Though we don’t yet know that $(0,1]$ is connected…

Before we get to that, let’s just go over a couple basic facts.  The image of a connected space under a continuous function is a connected space (for if not, the preimage of the separation would also be a separation.)  The product of spaces $X_\alpha$ is connected iff all $X_\alpha$ are.  (The “only if” follows from the first fact, since the projections are continuous.  For the “if” direction, assume you have a separation $A\cup B$ of $\prod X_\alpha$; since $p_\alpha(A)=p_\alpha(B)=X_\alpha$ for all but finitely many $\alpha$, it’s possible by process of elimination to find $a\in A,b\in B,\beta$ such that $p_\alpha(a)=p_\alpha(b)$ for all $\alpha\ne\beta$.  Then $p_\beta(A)\cup p_\beta(B)$ is a separation of $X_\beta$.)  The disjoint union, meanwhile, is disconnected if it’s of at least two spaces, and every connected component is contained in a “piece” of the disjoint union.  The union of connected subspaces with nonempty intersection is connected (which part of the separation contains the intersection?)  Most importantly, $\mathbb{R}$ is connected, as are all convex subsets.  We can prove this, but we need some terminology first.

Recall the definition of totally ordered set (toset).  A convex subset of a toset is one that includes the interval $[a,b]$ whenever $a$ and $b$ are in the subset.  Given a subset $A$ of a toset $T$, an upper bound of that subset is an element of $T$ that is greater than all elements of $A$.  A least upper bound is an upper bound that is less than all other upper bounds.  The least upper bound property is the property that all subsets with upper bounds have least upper bounds.  A linear continuum is a toset with the least upper bound property such that for all $a there is a $c$ with $a.  We know that $\mathbb{R}$ is totally ordered and has the second property above.  The least upper bound property actually follows from a construction we’ll later do (look up completeness if you’re curious), but for now I hope you’re willing to take it on faith.  So $\mathbb{R}$ is a linear continuum.

Theorem. Convex subsets of linear continua in the order topology are connected.  In particular, all intervals and rays of $\mathbb{R}$ are connected.

Proof.  Suppose we have a separation $C=A\cup B$ for $C$ a convex set. Pick points $a\in A,b\in B$, and assume WLOG that $a.  Then $[a,b]\subset C$, and we can let, say, $E=A\cap [a,b],F=B\cap [a,b]$, and again these should be a separation for $[a,b]$.  We’ll derive a contradiction to show that they are not actually a separation, so $A$ and $B$ are not a separation for $C$ either.

Observe first that $b$ is an upper bound for $E$, so $E$ has a least upper bound, say $c$.  This is in either $E$ or $F$.  If $F$, and $c\ne b$, then $c\in(d,e)\subset F$ since $F$ is open.  Since $c$ is an upper bound for $E$, the entire interval $(d,b]$ is in $F$; but then $d$ is a smaller upper bound for $E$!  If $c=b$, then you just get an interval $(d,c]$, but again $d is a contradiction.  If $c\in E$, though, we again get an interval $(d,e)$ or $[c,e)$ containing $c$ and inside $E$, and by the second property of linear continua, there’s a point between $c$ and $e$ that’s in $E$, so $c$ is not an upper bound.  Either way, we have a contradiction, so in fact there can be no separation.  $\square$

As a corollary, we get the good old Intermediate Value Theorem from calculus: any continuous function from $[a,b]\rightarrow [c,d]\subset\mathbb{R}$ takes on every value $e\in [c,d]$.  For if not, its image would have the separation $[a,e)\cup(e,d]$.  (I love it when more advanced math pumps out easy proofs of familiar theorems.)

But alas, all isn’t peaches and gravy.  We define a path in $X$ as a continuous map $\gamma:[0,1]\rightarrow X$.  Then $X$ is path-connected if for any $a,b\in X$, there is a path $\gamma$ with $\gamma(0)=a,\gamma(1)=b$.  The above discussion ensures that any path-connected space is connected, and all of the facts stated above carry over to the case of paths.

But the converse isn’t true.  The topologist’s sine curve, shown at right, is the graph of $y=\sin(1/x)$ with the subspace topology.  As it approaches $0$, it oscillates an infinite number of times between $1$ and $-1$.  Its closure is thus itself union the segment $[-1,1]$ on the $y$-axis.  Clearly the curve and the segment each lie in a whole connected component.  Is it the same connected component?  Yes, because connected components are clopen, hence closed, and the segment is part of the closure of the curve, so the only clopen is both of them.  On the other hand, no path connects a point on the curve to $(0,0)$.  A proof is that continuous functions send convergent sequences to convergent sequences, and every sequence converges in $[0,1]$, but we can find a sequence in the image that oscillates back and forth between $1$ and $-1$ an infinite number of times.

So the topologist’s sine curve is connected but not path connected.  We can restore the equality by using a local condition (every point has a (path-)connected basis of neighborhoods), which is something that probably merits its own mini-post, since I’m tired.  Just a note: outside of topology, path-connectedness seems far more natural and useful.  In differential geometry, you’re dealing with geodesics, which are “shortest paths,” and in algebraic topology, you’ve got path spaces for homotopy groups and singular 1-simplices for homology groups.  Then again, in nearly everything those subjects study, path-connectedness and connectedness are equivalent.  I’ll look out for counterexamples to my claim.  See you next time!

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You say that connected components are clopen, which is not true in general. It is true that they are closed, which follows from the fact that the closure of a connected subset is also connected and the maximality of connected components. But for example consider the set of rational points in R (subspace topology), the connected components are points (because Q are dense), and they are not open!

Comment by DMG