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Q&A Proving $|{\bf R}^{\bf R}|=|2^{\bf R}|$ using the Schroeder-Bernstein Theorem

2 answers  ·  posted 1y ago by Snoopy‭  ·  last activity 1y ago by celtschk‭

Question set-theory
#2: Post edited by user avatar trichoplax‭ · 2023-01-14T20:42:36Z (about 1 year 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 user avatar Snoopy‭ · 2023-01-14T14:38:38Z (about 1 year 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 $A$.