Explicit Injection of real-valued sequence space to real values
$begingroup$
Injection requires that for some mapping $f : mathbb{R}^mathbb{N} to mathbb{R}$, $f(x) = f(y) implies x = y$.
I managed to show the existence of one by using a series of equivalences:
- $mathbb{R}^mathbb{N} approx (2^mathbb{N})^mathbb{N} approx 2^{mathbb{N}^2}$
- $mathbb{N}^2 approx mathbb{N}$
- $2^mathbb{N} approx mathbb{R}$
So using the above facts we are able to see that there exists a bijection from $mathbb{R}^mathbb{N}$ to $mathbb{R}$.
However, is there an explicit mapping? I'm not sure if I'm allowed to use induction on uncountably many elements, otherwise I would construct it inductively.
real-analysis
asked Dec 12 '18 at 5:26
OneRaynyDayOneRaynyDay
306111
$endgroup$
add a comment |
$begingroup$
Injection requires that for some mapping $f : mathbb{R}^mathbb{N} to mathbb{R}$, $f(x) = f(y) implies x = y$.
I managed to show the existence of one by using a series of equivalences:
- $mathbb{R}^mathbb{N} approx (2^mathbb{N})^mathbb{N} approx 2^{mathbb{N}^2}$
- $mathbb{N}^2 approx mathbb{N}$
- $2^mathbb{N} approx mathbb{R}$
So using the above facts we are able to see that there exists a bijection from $mathbb{R}^mathbb{N}$ to $mathbb{R}$.
However, is there an explicit mapping? I'm not sure if I'm allowed to use induction on uncountably many elements, otherwise I would construct it inductively.
real-analysis
asked Dec 12 '18 at 5:26
OneRaynyDayOneRaynyDay
306111
$endgroup$
add a comment |
$begingroup$
Injection requires that for some mapping $f : mathbb{R}^mathbb{N} to mathbb{R}$, $f(x) = f(y) implies x = y$.
I managed to show the existence of one by using a series of equivalences:
- $mathbb{R}^mathbb{N} approx (2^mathbb{N})^mathbb{N} approx 2^{mathbb{N}^2}$
- $mathbb{N}^2 approx mathbb{N}$
- $2^mathbb{N} approx mathbb{R}$
So using the above facts we are able to see that there exists a bijection from $mathbb{R}^mathbb{N}$ to $mathbb{R}$.
However, is there an explicit mapping? I'm not sure if I'm allowed to use induction on uncountably many elements, otherwise I would construct it inductively.
real-analysis
asked Dec 12 '18 at 5:26
OneRaynyDayOneRaynyDay
306111
$endgroup$
Injection requires that for some mapping $f : mathbb{R}^mathbb{N} to mathbb{R}$, $f(x) = f(y) implies x = y$.
I managed to show the existence of one by using a series of equivalences:
- $mathbb{R}^mathbb{N} approx (2^mathbb{N})^mathbb{N} approx 2^{mathbb{N}^2}$
- $mathbb{N}^2 approx mathbb{N}$
- $2^mathbb{N} approx mathbb{R}$
So using the above facts we are able to see that there exists a bijection from $mathbb{R}^mathbb{N}$ to $mathbb{R}$.
However, is there an explicit mapping? I'm not sure if I'm allowed to use induction on uncountably many elements, otherwise I would construct it inductively.
real-analysis
real-analysis
asked Dec 12 '18 at 5:26
OneRaynyDayOneRaynyDay
306111
asked Dec 12 '18 at 5:26
OneRaynyDayOneRaynyDay
306111
asked Dec 12 '18 at 5:26
OneRaynyDayOneRaynyDay
306111
asked Dec 12 '18 at 5:26
OneRaynyDayOneRaynyDay
306111
asked Dec 12 '18 at 5:26
OneRaynyDayOneRaynyDay
306111
306111
add a comment |
add a comment |
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