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trap101
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Prove that the set i) S[itex]\subseteq[/itex] S(Closure)
ii) (Sint)int = Sint
Ok these supposedly simple containment questions irk me every time, how simplistic do I have to unravel the darn definitions:
Attempts: i) Let x [itex]\in[/itex] S(Closure)
==> x[itex]\in[/itex]S or x[itex]\in[/itex][itex]\partial[/itex]S (Boundary of S)
==> if x[itex]\in[/itex]S the there exists a B(r,x) [itex]\subseteq[/itex] S
likewise if x[itex]\in[/itex][itex]\partial[/itex]S then there exists the
B(r,x)[itex]\cap[/itex]S ≠∅ and B(r,x)[itex]\cap[/itex]Sc≠∅
==> S[itex]\subseteq[/itex] S(Closure) I really don't see what else can be done
ii) isn't this just direct from the definition of the Sint? I mean it's the interior of an interior.
ii) (Sint)int = Sint
Ok these supposedly simple containment questions irk me every time, how simplistic do I have to unravel the darn definitions:
Attempts: i) Let x [itex]\in[/itex] S(Closure)
==> x[itex]\in[/itex]S or x[itex]\in[/itex][itex]\partial[/itex]S (Boundary of S)
==> if x[itex]\in[/itex]S the there exists a B(r,x) [itex]\subseteq[/itex] S
likewise if x[itex]\in[/itex][itex]\partial[/itex]S then there exists the
B(r,x)[itex]\cap[/itex]S ≠∅ and B(r,x)[itex]\cap[/itex]Sc≠∅
==> S[itex]\subseteq[/itex] S(Closure) I really don't see what else can be done
ii) isn't this just direct from the definition of the Sint? I mean it's the interior of an interior.