Understanding Countability of Set L: A Confusing Point

[Bachelier]

I don't understand this point:

Given the open set E = U_(a in L) I_a. Union of open intervals
We're showing this is countable.

WTS is that indexed set L is countable.

Set g: L---> Q (rationals) because Q is dense then every interval meets Q.
a---> q_a

this is 1-1. But here's where I get confused:

The conclusion is: Hence L is bijectively equivalent to a subset of Q and hence is countab.

I understand the fact that a subset of countable set is countable. But g codomain is Q. Should I restrict it to some Q_a in Q and show that g is invertible. Hence bij.
We only proved g is 1-1. Where did the bij. argument come from.

Thanks guys

 

[chiro]

This is just a guess, but a bijection from a countable set to a uncountable set can still be bijective but not onto.

Im guessing that you if you show that the function is not onto, then with the property of the bijection, then the range of the function is countable.

 

[Bachelier]

This is just a guess, but a bijection from a countable set to a uncountable set can still be bijective but not onto.

Im guessing that you if you show that the function is not onto, then with the property of the bijection, then the range of the function is countable.

Yeah but the only thing known about the function "g" is that it is injective.
It sure is not surjective on Q although this latter is countable.

 

[Bachelier]

any ideas?

 

[chiro]

I haven't done enough pure math to formulate what I mean precisely in mathematical language but intuitively if you have a 1-1 between domain and range then if domain is countable then range is countable since if you have finite domain you have finite range.

Surely there is a way of saying if you have finite domain and mapping is 1-1 then range must be finite. Only things I can think of is using cardinality or maybe the union of the sets in the range (something like if f(A) = B then union of range is union f(A) over A which implies union (B) over rationals which is subset of rationals (prove that all function values are rational) and since each is disjoint (because of 1-1) the size of the set is the number of elements which is size of L).

I'm sorry I can't put it into the "theorem/proof" kind of way but it is a fairly intuitive result that a finite mapping that is 1-1 must have a finite range.