Let S be a compact subset of a hausdorff space X. Prove that the boundary of S is compact.
Let S be a subset of a hausdorff space X and let B be it's boundary. Let x be a limit point of B. Then every neighbourhood of x contains an element y in B different to x. So there exists disjoint open sets U,V such that x is in U, y is in V.
Now i'm stuck.
Let's use $\displaystyle \beta(S)$ for the boundary of $\displaystyle S$.
What is the exact meaning of $\displaystyle t\in\beta(S)$. (i.e. what is a boundary point?)
Note how I edited your reply. Say $\displaystyle U$ is a neighborhood of x.
Then there is a neighborhood of y, $\displaystyle V$, such that $\displaystyle x\not\in V\subset U.$.
Now apply the definition of a boundary point.
That is your whole problem in a nutshell.
Let this be a lesson to you. Always post what you know about the question. I had no way of knowing that you were working with what I consider a non-standard definition.
Notation: $\displaystyle \mathcal{I}(S)$ stands for the interior of $\displaystyle S$.
By your definition $\displaystyle \beta(S)=\overline{S}\setminus\mathcal{I}(S)$.
But $\displaystyle \mathcal{I}(S)$ is open set. Therefore its complement is closed.
Thus $\displaystyle \beta(S)$ is the intersection of two closed sets. SO?
Are you sure you've proved the boundary is compact, and not closed? Something's wrong because it is a necessary hypothesis
BTW: are you highlighting the capital or that it should be droff and not dorff?
Thanks for your help. In future please don't respond to my questions as other (better) helpers will see you're dealing with it and I will be stuck with your minimalist answers. I know your ethos is let the student do it themselves but sometimes it is more instructive for me to see the proof.