Question

this is a question in topology, prove that if A is a countable dense set of IR then there exists a subset B of A dense and countable as well.

Answer 1

hey :D

Answer 2

so would you mind to write countable dense def ?
tbh this proof is rigid xD

Answer 3

Hi, I'll start first with def of a dense set we say that A is dense if $\bar{A}=\mathbb{R}$

Answer 4

yes sorry

Answer 5

u ment this , right ?

\( \bar{A}=\mathbb{R} \)

Answer 6

\[\bar{A}=\mathbb{R}\]

Answer 7

ohk :D continue

Answer 8

and A is countable if there is an injective function from A to \[\mathbb{N}\]

Answer 9

well do you know that
every infinite set has a countable dense subset ?

Answer 10

no actually that is a part of what we're trying to prove , no?

Answer 11

well i was trying to reach the proof by showing that there exist A infinite group ...
mmm but since you dnt know it , lets try something else

Answer 12

okay

Answer 13

so since A is a countable dense set of IR
then
1-\(\bar{A}=\mathbb{R}\)
2- there is F(x)=y , s.t f(x) is 1_1 (injective )

i still cant catch a line -.-

i mean what i know is each subset of sense set is also dense

also , if a set have 1_1 injective function then there exist a subset that also have 1_1 injective ..


im lost with this , sry xD

Answer 14

it's okay that made me confused too :p

Answer 15

:D