# Is a convex subset of a connected space connected?

1. Jan 31, 2012

### nonequilibrium

It seems like something that could (should?) be true, but with topology you never know (unless you prove it...).

EDIT: I'll be more exact: let $(X,\mathcal T)$ be a topological space with X a totally ordered set and $\mathcal T$ the order topology. Say X is connected and $A \subset X$ is convex (i.e. $\forall a,b \in A: a < b \Rightarrow [a,b] \subset A$). Is A connected?

2. Jan 31, 2012

### micromass

Let X be an ordered set equiped with the order topology. If X is connected, then you can show that

1) Every set that is bounded from above has a least upper bound
2) If x<y, then there is a z such that x<z<y.

Try to prove this.

Then you can prove that all intervals are connected just as you prove this with $\mathbb{R}$.

3. Jan 31, 2012

### nonequilibrium

For future reference:

1) Let A be a subset of X bounded from above but without least upper bound, then define the non-trivial subset B as the set of all upper bounds.
B is closed: take a $b \in \overline B$, and suppose there is an $a \in A$ with b < a, then $]- \infty, a[$ is an open around b that does not intersect B. Contradiction, hence b is an upper bound of A.
B is open: take a $b\in B$, since b is not the smallest upper bound, there is a $\beta \in B$ with $\beta < b$, hence $b \in ]\beta,+ \infty[ \subset B$.

2) Quite obvious indeed :) otherwise one can construct a separation of X.

By golly you're right. As you probably know, the converse is a theorem in Munkres (section 24), but I hadn't realized it was a characterization. So every connected order topology corresponds to a linear continuum. Nice to know! Thanks :)