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I'm taking second year Calc in my university, and we were told to prove this:

Let $S$ be an open set in $\mathbb{R}^n$ and $p \in S$ and $q \notin S$. Prove that a boundary point of $S$ is on the line segment joining $p$ and $q$.

I know that it's obvious, but can't seem to actually prove this result. I'm in a somewhat basic course, so I know a few things: connectedness and disconnectedness, how to use balls in an open set, what an accumulation point is, but some of the more rigorous topological terms might be unfamiliar.

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    @ChrisEagle: I would say $x$ is a boundary point of $S$ if every open neighborhood of $x$ intersects both $S$ and the complement of $S$.2011-10-23

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Hint: Consider the real number $t^*=\sup\{t\in[0,1]\mid tq+(1-t)p\in S\}$ and the point $p^*=t^*q+(1-t^*)p$.

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    Right. Likewise, if $(p^*-\varepsilon,p^*)\cap S=\varnothing$, then... Once you feel you have completed a proof, THE thing to do is to post the result here as an answer so we can check that it is correct.2011-10-24