Definition of Connectedness

**JG89** · March 6th 2010, 04:39 PM

In my book, "Real Mathematical Analysis" by Charles Pugh, a subset S of a metric space M is disconnected if there exists a proper, clopen subset of Because if the proper, clopen subset of S is A, then S = A union A(compliment)

But for example on this wikipedia page: Connectedness - Wikipedia, the free encyclopedia

It says that S is not connected (so disconnected) if it is not the union of two disjoint non-empty open sets, which is basically saying that S has no proper, open set.

Which is the correct definition?

**Plato** · March 6th 2010, 04:52 PM

Originally Posted by JG89

In my book, "Real Mathematical Analysis" by Charles Pugh, a subset S of a metric space M is disconnected if there exists a proper, clopen subset of Because if the proper, clopen subset of S is A, then S = A union A(compliment)
But for example on this wikipedia page: Connectedness - Wikipedia, the free encyclopedia
It says that S is not connected (so disconnected) if it is not the union of two disjoint non-empty open sets, which is basically saying that S has no proper, open set.

To confuse you even more here is a third definition given by R L Moore in c1920.
Two nonempty sets are separated if neither one contains a point nor a limit point of the other.
(Of course, Prof Moore did not say nonempty- he did not believe empty point sets existed.)
A set is connected if and only if it is not the union of two separated sets.

But here is the kicker: All three are equivalent. You ought to prove it.

**JG89** · March 6th 2010, 07:37 PM

Thanks! I'll get on that.

**HallsofIvy** · March 7th 2010, 04:43 AM

Be careful, they are not exactly equivalent. The definition by Moore, quoted by Plato, is for connected sets. The other two definitions are for connected spaces. For example, $A= [0, 1]\cup [2, 3]$ is not a connected set in R (with the "usual" topology), but the only "clopen" sets in R are the empty set and R itself.

Of course, if you think of A, above, as a vector space in its own right, with the topology inherited from R, then [0,1] and [2,3] are "clopen" in A.

**JG89** · March 7th 2010, 07:58 AM

Halls, if I am studying metric spaces, should I use the clopen and open set definitions?

**Plato** · March 7th 2010, 09:22 AM

Originally Posted by JG89

I am studying metric spaces, should I use the clopen and open set definitions?

It is save to say that in many textbooks where metric spaces are the central focus, a space is said to be connected if it is not the union of two nonempty disjoint open sets. If a metric space is not connected, then the two separating sets are both open and closed (clopen).

**JG89** · March 9th 2010, 02:02 PM

Originally Posted by Plato

You ought to prove it.

Definition 1: A metric space $M$ is connected if there is not a proper, clopen subset of M.

Definition 2: A metric space $M$ is connected if $M$ is not the union, of two disjoint, non-empty open sets.

Proof that the two definitions are equivalent:

First suppose that $M$ is not the union of two disjoint, non-empty open sets. We prove that it does not contain a proper, clopen subset. Assume that indeed it does, $A \subset M$ is clopen. Then $M = A \cup A^c$ and $A \cap A^c = \varnothing$ . Both $A$ and $A^c$ are clopen, and thus open, sets that are disjoint and non-empty (we know they are non-empty because they are assumed to be proper subsets), which contradicts the fact that M is not the union of two disjoint, non-empty open sets.

Now we must prove the other direction. We will prove it in contrapositive form. Suppose that $M$ is the union of two disjoint, non-empty open sets, $A$ and $B$ . We will prove that one of them is clopen. Take either $A$ or $B$ , say $A$ . Suppose it is not closed. There exists a convergent sequence $a_n \rightarrow a$ of its elements whose limit is not in $A$ , and thus must be in $B$ . Remember that $B$ is open and so $\exists \delta > 0 : d(a, x) < \delta \Rightarrow x \in B$ . Note that there exists at least one point in $\{x \in M : d(a,x) < \delta \}$ that's from the sequence $a_n$ since $a_n \rightarrow a$ . But this contradicts the fact that $A \cap B = \varnothing$ , and so $A$ is closed, and since it is also open, it is clopen. $A$ is also a proper subset, and so there exists a proper, clopen subset of $M$ . QED

Is this fine?

**Plato** · March 9th 2010, 02:14 PM

That proof works.
In the first part you need to be sure that set A is a proper subset.

As for the second part recall that the complement of an open set is closed.
Therefore $M\setminus A=B$ so the set $B$ is both open and closed.

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