1. Prove that the set of all irrational numbers is dense in R. Hint: R is seperable so there exists a sequence of open sets that form a base. Use sequences, show that sequence is dense in R, therefore conclude set of irrationals is dense in R

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(I included page from book that was said to follow idea as guide Theorem 5.10)

Thankyou, I dont know how prove using sequences

1. Prove that the set of all irrational numbers is dense in R.
Hint: R is seperable so there exists a sequence of open sets that form a base. Use sequences,
show that sequence is dense in R, therefore conclude set of irrationals is dense in R
Transcribed Image Text:1. Prove that the set of all irrational numbers is dense in R. Hint: R is seperable so there exists a sequence of open sets that form a base. Use sequences, show that sequence is dense in R, therefore conclude set of irrationals is dense in R
union or
roIC oboce M
noiam
5.10 Theorem: A metric space is second-countable if and only if it is separable
Proof: Suppose first that X is a separable metric space. Let {x,}j=1 be a dense 5.12
sequence in X. Consider the family of open sets
bns ti
bos Yo
B
{B(x;1/n) : j > 1, n z 1}.
%3D
EXE
Let U be an open subset of X and let x e U. For some n 2 1, we have B(x;2/n) C
U. Choose j so that d(xjx) < 1/n. Then x e he
shows that B(x;1/n) C B(x;2/n) C U. Consequently B is a base of open sets. Since
B is countable, X is second-countable.
Conversely, suppose that X is second-countable. Let {U-1 be a sequence of
open sets in X that form a base. Let x, be any point in U,, n z 1. Then every
nonempty open subset of X contains a point of the sequence {x}, so that the sequence
is dense in X, and X is separable. O
€ B(x;1/n), and the triangle inequality
The following theorem is the trickiest part of the circle of ideas covered in this
section.
1gaiai doso bns Xax doso noi.Xlo isadue oldsin
Transcribed Image Text:union or roIC oboce M noiam 5.10 Theorem: A metric space is second-countable if and only if it is separable Proof: Suppose first that X is a separable metric space. Let {x,}j=1 be a dense 5.12 sequence in X. Consider the family of open sets bns ti bos Yo B {B(x;1/n) : j > 1, n z 1}. %3D EXE Let U be an open subset of X and let x e U. For some n 2 1, we have B(x;2/n) C U. Choose j so that d(xjx) < 1/n. Then x e he shows that B(x;1/n) C B(x;2/n) C U. Consequently B is a base of open sets. Since B is countable, X is second-countable. Conversely, suppose that X is second-countable. Let {U-1 be a sequence of open sets in X that form a base. Let x, be any point in U,, n z 1. Then every nonempty open subset of X contains a point of the sequence {x}, so that the sequence is dense in X, and X is separable. O € B(x;1/n), and the triangle inequality The following theorem is the trickiest part of the circle of ideas covered in this section. 1gaiai doso bns Xax doso noi.Xlo isadue oldsin
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