Show an open set has no minimum Suppose $A$ is an open subset of the real numbers. How to prove $A$ has no minimum?
I guess it should be done with Proof-by-contradiction, but I don't get anywhere at this point.

Suppose $m \in A$ is the minimum of A. By definition of open set, you have that $(m-\delta , m + \delta) \subseteq A$ with $\delta > 0$ sufficiently small, then $m- \frac{\delta}{2} \in A$ but this is a contradiction (because we have found an element in A smaller than $m$)
