Difference between revisions of "Schroeder-Bernstein Theorem"
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===Introductory=== | ===Introductory=== | ||
+ | ====Problem 1==== | ||
Show that <math>\mathbb{Q}</math> is countable. | Show that <math>\mathbb{Q}</math> is countable. | ||
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+ | ====Problem 2==== | ||
+ | Let <math>\kappa</math> satisfy <math>\kappa\cdot\kappa=\kappa</math>. Show that <math>\kappa^{\kappa}=2^{\kappa}</math>. | ||
===Intermediate=== | ===Intermediate=== | ||
====Problem 1==== | ====Problem 1==== | ||
− | We say a set of reals <math>O\subseteq\mathbb{R}</math> is | + | We say a set of reals <math>O\subseteq\mathbb{R}</math> is open if for all <math>r\in O</math>, there is an open interval <math>I</math> satisfying <math>r\in I\subseteq O</math>. Show that the following sets are equal in cardinality: |
*<math>\mathbb{R}</math> | *<math>\mathbb{R}</math> | ||
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*<math>\{O\subset\mathbb{R}\mid O\text{ is open}\}</math> | *<math>\{O\subset\mathbb{R}\mid O\text{ is open}\}</math> | ||
*<math>\{f:\mathbb{R}\rightarrow\mathbb{R}\mid f\text{ is continuous}\}</math> | *<math>\{f:\mathbb{R}\rightarrow\mathbb{R}\mid f\text{ is continuous}\}</math> | ||
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==See Also== | ==See Also== | ||
[[Category:Set theory]] | [[Category:Set theory]] | ||
[[Category:Theorems]] | [[Category:Theorems]] |
Latest revision as of 12:09, 20 February 2024
The Schroeder-Bernstein Theorem (sometimes called the Cantor-Schroeder-Bernstein Theorem) is a result from set theory, named for Ernst Schröder and Felix Bernstein. Informally, it implies that if two cardinalities are both less than or equal to each other, then they are equal.
More specifically, the theorem states that if and are sets, and there are injections and , then there is a bijection .
The proof of the theorem does not depend on the axiom of choice, but only on the classical Zermelo-Fraenkel axioms.
Contents
[hide]Proof
We call an element of lonely if there is no element such that . We say that an element of is a descendent of an element of if there is a natural number (possibly zero) such that .
We define the function as follows: Note that cannot be lonely itself. If is the descendent of a lonely point, then for some ; since is injective, the element is well defined. Thus our function is well defined. We claim that it is a bijection from to .
We first prove that is surjective. Indeed, if is the descendent of a lonely point, then ; and if is not the descendent of a lonely point, then is not lonely, so there is some such that ; by our definition, then, . Thus is surjective.
Next, we prove that is injective. We first note that for any , the point is a descendent of a lonely point if and only if is a descendent of a lonely point. Now suppose that we have two elements such that . We consider two cases.
If is the descendent of a lonely point, then so is . Then . Since is a well defined function, it follows that .
On the other hand, if is not a descendent of a lonely point, then neither is . It follows that . Since is injective, .
Thus is injective. Since is surjective and injective, it is bijective, as desired.
Problems
The Schroeder-Bernstein Theorem can be used to solve many cardinal arithmetic problems. For example, one may wish to show for some cardinal . One strategy is to find sets such that with injections from to and to , thus concluding that . More generally, the Schroeder-Bernstein Theorem shows that the relation between cardinals and defined by "there is an injection " is a partial-order on the class of cardinals.
Introductory
Problem 1
Show that is countable.
Problem 2
Let satisfy . Show that .
Intermediate
Problem 1
We say a set of reals is open if for all , there is an open interval satisfying . Show that the following sets are equal in cardinality: