Connectedness of a subset of a metric space.

[tylerc1991]

Homework Statement

Show that an open (closed) subset of a metric space E is connected if and only if it is not the disjoint union of two nonempty open (closed) subsets of E.

Homework Equations

The definition of connectedness that we are using is as follows:
A metric space E is connected if the only subsets of E which are both open and closed are E and $\varnothing$

The Attempt at a Solution

The first issue I had with the problem is the wording. Do they mean that there are two cases? one with open sets and one with closed sets? Anyway, I'll proceed the way that I interpreted the question, and you can stop me if I missed what they were asking.

I'll try the 'open' part first. That is, I will try to show that an open subset of a metric space E is connected if and only if it is not the disjoint union of two nonempty open subsets of E. The proof for the 'closed' part should be similar, so I will omit that here.

This statement is equivalent to showing that an open subset of a metric space E is *not* connected if and only if it *is* the disjoint union of two nonempty open subsets of E.

I'll start with the 'if' direction first. First, let S be an open subset of some metric space E. Suppose that S is the disjoint union of two nonempty open subsets of E, call them A and B. That is, $S = A \cup B$, $A \cap B = \varnothing$, A and B are nonempty (which implies $A \neq S$ and $B \neq S$), and A and B are open. Now comes the part that I felt uneasy about: Since S is the space under consideration, and A and B union to S, I may say that $B = A^{c}$ and $A = B^{c}$. After this is easy, because I have found a set (actually two) that are both open and closed (because the complement of an open set is closed and vice versa). By the definition of connectedness above, this means that S is not connected, which is what I am after.

Now for the 'only if' part. Again let S be an open subset of a metric space E. Suppose that S is not connected. By the definition above, there exists a set A such that $A \neq \varnothing$, $A \neq S$, and A is both open and closed. Again the uneasy part: Let $B = A^{c}$. Then $A \cup B = S$, $A \cap B = \varnothing$, and A and B are both open (and closed for that matter), completing the other direction.

Please let me know if I am justified in writing the uneasy parts, and if I am not, please help me fix them! Thank you for your time!

 

[mighty2000]

Your proof looks correct to me! The only thing I would suggest is to maybe add a little more explanation for why you can say that B=A^c and A=B^c. This is because if A is open and closed, then its complement B is also open and closed, and since S is open, its complement must be closed. So we can say that B=A^c and A=B^c because they are complements of each other and both open and closed. Other than that, great job!