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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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    What is your definition of "boundary point"?2011-10-23
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    If not, the segment $[p,q]$ would be partitioned by the two sets of points respectively interior and exterior to $S$. What's insufferable about that?2011-10-23
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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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    I'm not actually sure how to make use of this. It seems to define the boundary point in some manner, but I'm not sure what it's doing.2011-10-23
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    Does the number define a function that maps the line to a connected set and then defines a point on that line that would be on the boundary and translate into a point on the boundary?2011-10-23
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    Not sure I follow everything you say but yes, the function $t\mapsto tq+(1-t)p$ does map continuously the interval $[0,1]$ to a subset of $\mathbb R^n$ and the idea of the hint is that the point $p^*$ could be on the boundary of $S$. To show this, one could try to show that some points infinitely close to $p^*$ are in $S$ and some others are not. But first, one has to get rid of the case $t^*=1$. Can you do that? Second, can you show that $t^*=0$ is impossible? And third, can you deal with the case when $t^*$ is in $(0,1)$?2011-10-23
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    $t^* = 1$ implies q is on the boundary and we're done. $t^* = 0$ implies $q \in S$, so we have a problem.2011-10-23
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    Right on both counts. Hence you are left with the case when $0$p^*$ is on the boundary? – 2011-10-23
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    I would say that any neighborhood around $p^*$ would contain points both in $S$ and in $S^c$, but I don't know how to show it.2011-10-23
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    Try to show that $(p^*-\varepsilon,p^*)\cap S\ne\varnothing$ for every positive $\varepsilon$ and that $p^*$ is not in $S$.2011-10-23
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    I'm really just not getting it.2011-10-24
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    Let us begin with the proof that $p^*$ is not in $S$. If $p^*$ is in $S$, since $S$ is open, a ball around $p^*$ is included in $S$. That means that for every $|t-t^*|\leqslant\varepsilon$, $tq+(1-t)p$ is in $S$. Hence...2011-10-24
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    there's a contradiction with the supremum because we have a higher bound.2011-10-24
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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