- #1
Mr Davis 97
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Homework Statement
Suppose that S is a closed set. We claim that Sc is open. Take any p ∈ Sc. If
there fails to exist an r > 0 such that
d(p, q) < r ⇒ q ∈ Sc
then for each r = 1/n with n = 1, 2, . . . there exists a point pn ∈ S such that
d(p, pn) < 1/n. This sequence in S converges to p ∈ Sc, contrary to closedness of S.
Therefore there actually does exist an r > 0 such that
d(p, q) < r ⇒ q ∈ Sc
which proves that Sc is an open set.
Homework Equations
The Attempt at a Solution
So my question here is not how to do the prove, because I have that proof right here. I am just confused about the proof. I understand that the argument in general is a proof by contradiction, by assuming that Sc is not open and deriving a contradiction. However, I am confused about the reasoning here:
If there fails to exist an r > 0 such that
d(p, q) < r ⇒ q ∈ Sc
then for each r = 1/n with n = 1, 2, . . . there exists a point pn ∈ S such that
d(p, pn) < 1/n. This sequence in S converges to p ∈ Sc.
Can someone explain the logic here? Where does the 1/n comes from? Why does that sequence converge to p?