A and A complement have the same boundary

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The discussion centers on proving that if X is a non-empty set and A is a subset of X, then the boundary of A, denoted as ∂A, is identical to the boundary of its complement, A^c. The proof relies on the definition of the boundary, where a point x belongs to ∂A if there exists a radius r > 0 such that the open ball B(x, r) intersects both A and A^c. This definition directly implies that x also belongs to ∂A^c, establishing that ∂A = ∂A^c as a straightforward conclusion.

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How would you show that if [itex]X[/itex] is a non-empty set and [itex]A\subseteq X[/itex] then [itex]A[/itex] and [itex]A^c[/itex] have the same boundary?

The definition is [itex]x\in \partial A \iff[/itex] there exists [itex]r>0[/itex] such that the open ball [itex]B(x,r)[/itex] intersects both [itex]A[/itex] and [itex]A^c[/itex]

but this is precisely the statement that [itex]x\in \partial A^c[/itex]!
 
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Ted123 said:
How would you show that if [itex]X[/itex] is a non-empty set and [itex]A\subseteq X[/itex] then [itex]A[/itex] and [itex]A^c[/itex] have the same boundary?

The definition is [itex]x\in \partial A \iff[/itex] there exists [itex]r>0[/itex] such that the open ball [itex]B(x,r)[/itex] intersects both [itex]A[/itex] and [itex]A^c[/itex]

but this is precisely the statement that [itex]x\in \partial A^c[/itex]!

Indeed, so the proof is trivial.
 

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