Showing that one open set contains a compact set does not prove what the OP wanted!
However, I can finish by showing the "easy" part: If A is not empty, then it contains, say, "a". The singleton set "a", since it is finite, is compact.
#4
g_edgar
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My point was, that every open set contains the empty open set, which contains the empty compact set. So I was not really doing *one* open set after all.
I'm reviewing Meirovitch's "Methods of Analytical Dynamics," and I don't understand the commutation of the derivative from r to dr:
$$
\mathbf{F} \cdot d\mathbf{r} = m \ddot{\mathbf{r}} \cdot d\mathbf{r} = m\mathbf{\dot{r}} \cdot d\mathbf{\dot{r}}
$$