[Math] Can an open set contain its supremum

general-topologyreal-analysis

I say the answer is no:

By definition,let $A \subset \mathbb{R} $ and let $s$ be an upperbound for $A$. Then, $s=sup A$ if and only if, $\forall \epsilon>0$, $\exists a \in A$ satisfying $s-\epsilon < a$

With this in mind, and I'm having trouble stating it formally, let's assume that $A \subset \mathbb{R}$ is open, as the question states. For a contradiction, let's say that $s \in A$. Let $b$ be an upperbound for $A$. Since $s = sup A$, for an arbitrary $\epsilon$, $s + \epsilon \leq b$, but $b \notin A$.

Where I am going with this is that $V_\epsilon (s)$ is not contained in $A$, which contradicts the fact that $A$ is an open set.

How is my logic and how can I improve on that?
Thanks!

Best Answer

In the reals, using the Euclidean topology, you are right and your argument works. I would phrase it as:

Let $s$ be an upper bound for $A$. Then for some $\epsilon$, there is a ball $B_{\epsilon}(s)$ around $s$ contained in $A$, since $A$ is open. But in particular, $s+\frac{\epsilon}{2}$ lies in $A$, contradicting that $s$ is an upper bound.

If you are averse to contradiction, you can do it directly: for every $s \in A$, there is $s+\epsilon/2 \in A$ with $s+\epsilon/2 > s$.

In a general partially ordered topological space, though, open sets may contain their suprema: in $[0,1]$ with the subspace Euclidean topology, $[0,1]$ is open and contains its sup, $1$.

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