1. Definition of Connectedness

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?

2. 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.

3. Thanks! I'll get on that.

4. 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.

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

6. 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).

7. 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?

8. 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.