For a set X with the discrete topology, show that for every $A\subset X$:

  1. $\text{int} A = A$
  2. $\text{ext} A = X\setminus A$
  3. $\partial A = \emptyset$

where int means the interior of $A$, ext means the exterior of $A$, and $\partial A$ is the boundary of $A$.

Well, I try by the open sets but I got nowhere... Then I try by neighborhood and I got nowhere...

Of course that the proof of the boundary of A is really obvious, but I can't do the int....can somebody help?

  • $\begingroup$ Maybe you could add to your question what you consider the definition of $\operatorname{int} A$. (This might help user answering your question to provide an answer which suits your needs.) $\endgroup$ – Martin Sleziak Dec 11 '15 at 4:09

The interior of a set S is the set of all points of S that are in some open set contained in S. But every set is open (in the discrete topology), so if x is in S, then the set { x } is an open set contained in S.

  • $\begingroup$ But how do you prove it by neighborhood? $\endgroup$ – pipita Jul 21 '15 at 16:42
  • $\begingroup$ I'm not sure I understand the question. For the first part, you just have to show that every x in A is an interior point, you can do so by finding an open set that x is in (or a "neighborhood of x", that means the same thing), that is completely contained in A. $\endgroup$ – Athar Abdul-Quader Jul 21 '15 at 16:53

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