Open/closed set and interior point problem

  • #1

Homework Statement



Let [tex](X,d)[/tex] be a metric space and let [tex]A \subseteq X[/tex]. Denote the interior of [tex]A[/tex] by [tex]A^o[/tex].

Homework Equations



Prove that if [tex]A[/tex] is open or closed, then [tex](\partial A)^o = \varnothing[/tex]. (Is this still true if [tex]A[/tex] is not open or closed?)

The Attempt at a Solution



I don't even know what [tex]\partial A[/tex] means. Can anyone tell me what this represents?
 

Answers and Replies

  • #3
That notation denotes the boundary of A.
http://en.wikipedia.org/wiki/Boundary_(topology)

[tex]
\begin{align*}
\partial A = \overline{A} \cap \overline{X - A} = \{ x \in
X | \forall \varepsilon > 0 (O_\varepsilon(x) \cap A \ne \varnothing
\wedge O_\varepsilon(x) \cap (X-A) \ne \varnothing \}
\end{align*}
[/tex]

If [tex]A[/tex] is closed, then [tex]\partial A \subseteq A[/tex]. However, [tex]\forall x
\in \partial A \forall \varepsilon > 0 ( O_\varepsilon(x) \cap (X -
A) \ne \varnothing)[/tex], hence [tex]\forall x \in
\partial A \nexists \varepsilon > 0 (O_\varepsilon(x) \subseteq
\partial A)[/tex]. There are no interior points.

If [tex]A[/tex] is open, then [tex]\partial A \subseteq X - A[/tex]. However, [tex]\forall
x \in \partial A \forall \varepsilon > 0 ( O_\varepsilon(x) \cap A
\ne \varnothing)[/tex], hence [tex]\forall x \in
\partial A \nexists \varepsilon > 0 (O_\varepsilon(x) \subseteq
\partial A)[/tex]. There are no interior points.

Is this correct? Is it still true if [tex]A[/tex] is not open or closed? How can I prove that?
 
  • #4
jbunniii
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Is it still true if [tex]A[/tex] is not open or closed? How can I prove that?

Consider [itex]A = \mathbb{Q}[/itex], [itex]X = \mathbb{R}[/itex]. What is the boundary of [itex]\mathbb{Q}[/itex]?
 

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