Post History
#2: Post edited
by
trichoplax
·
2023-01-14T20:42:36Z (almost 2 years ago)
Typo
Proving $|{\bf R}^{\bf R}|=|2^{\bf R}|$ using the Schroeder-Bernstein Theorem
- Let $A$ be the set of *all* functions from ${\bf R}$ to ${\bf R}$ and $B$ the power set of ${\bf R}$. Then $|A|=|B|$.
- This is a well-known result in set theory. A quick search on Google returns answers in various places explicitly using [cardinal arithmetic](https://en.wikipedia.org/wiki/Cardinal_number#Cardinal_arithmetic).
- **Question**: Can one prove the above result _using_ the [Schroeder-Bernstein Theorem](https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem)?
- ---
- Remark.
To use Schroeder-Bersnstein, one needs two injections. One easy direction is from $B$ to $A$: any subset of ${\bf R}$ can be identifies as a function $f:{\bf R}\to\{0,1\}$. Such identification gives an injection from $B$ to $A$.
- Let $A$ be the set of *all* functions from ${\bf R}$ to ${\bf R}$ and $B$ the power set of ${\bf R}$. Then $|A|=|B|$.
- This is a well-known result in set theory. A quick search on Google returns answers in various places explicitly using [cardinal arithmetic](https://en.wikipedia.org/wiki/Cardinal_number#Cardinal_arithmetic).
- **Question**: Can one prove the above result _using_ the [Schroeder-Bernstein Theorem](https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem)?
- ---
- Remark.
- To use Schroeder-Bersnstein, one needs two injections. One easy direction is from $B$ to $A$: any subset of ${\bf R}$ can be identified as a function $f:{\bf R}\to\{0,1\}$. Such identification gives an injection from $B$ to $A$.
#1: Initial revision
by
Snoopy
·
2023-01-14T14:38:38Z (almost 2 years ago)
Proving $|{\bf R}^{\bf R}|=|2^{\bf R}|$ using the Schroeder-Bernstein Theorem
Let $A$ be the set of *all* functions from ${\bf R}$ to ${\bf R}$ and $B$ the power set of ${\bf R}$. Then $|A|=|B|$.
This is a well-known result in set theory. A quick search on Google returns answers in various places explicitly using [cardinal arithmetic](https://en.wikipedia.org/wiki/Cardinal_number#Cardinal_arithmetic).
**Question**: Can one prove the above result _using_ the [Schroeder-Bernstein Theorem](https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem)?
---
Remark.
To use Schroeder-Bersnstein, one needs two injections. One easy direction is from $B$ to $A$: any subset of ${\bf R}$ can be identifies as a function $f:{\bf R}\to\{0,1\}$. Such identification gives an injection from $B$ to $A$.