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Difference between revisions of "Schroeder-Bernstein Theorem"

m (Added clarification that $f(a)$ cannot be lonely; which is why f(a) = f(g(b)) for some b.)
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Thus <math>h</math> is injective.  Since <math>h</math> is surjective and injective, it is
 
Thus <math>h</math> is injective.  Since <math>h</math> is surjective and injective, it is
 
bijective, as desired.  <math>\blacksquare</math>
 
bijective, as desired.  <math>\blacksquare</math>
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==Problems==
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The Schroeder-Bernstein Theorem can be used to solve many cardinal arithmetic problems. For example, one may wish to show <math>|S|=\kappa</math> for some cardinal <math>\kappa</math>. One strategy is to find sets <math>A,B</math> such that <math>|A|=|B|=\kappa</math> with injections from <math>A</math> to <math>S</math> and <math>S</math> to <math>B</math>, thus concluding that <math>|A|=|S|=|B|=\kappa</math>. More generally, the Schroeder-Bernstein Theorem shows that the relation <math>|A|\leq|B|</math> between cardinals <math>|A|</math> and <math>|B|</math> defined by "there is an injection <math>f:A\rightarrow B</math>" is a partial-order on the class of cardinals.
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===Introductory===
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====Problem 1====
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Show that <math>\mathbb{Q}</math> is countable.
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====Problem 2====
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Let <math>\kappa</math> satisfy <math>\kappa\cdot\kappa=\kappa</math>. Show that <math>\kappa^{\kappa}=2^{\kappa}</math>.
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===Intermediate===
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====Problem 1====
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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:
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*<math>\mathbb{R}</math>
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*<math>2^{\aleph_{0}}</math>
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*<math>\{O\subset\mathbb{R}\mid O\text{ is open}\}</math>
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*<math>\{f:\mathbb{R}\rightarrow\mathbb{R}\mid f\text{ is continuous}\}</math>
  
 
==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 $A$ and $B$ are sets, and there are injections $f: A \to B$ and $g : B \to A$, then there is a bijection $h : A \to B$.

The proof of the theorem does not depend on the axiom of choice, but only on the classical Zermelo-Fraenkel axioms.

Proof

We call an element $b$ of $B$ lonely if there is no element $a \in A$ such that $f(a) = b$. We say that an element $b_1$ of $B$ is a descendent of an element $b_0$ of $B$ if there is a natural number $n$ (possibly zero) such that $b_1 = (f \circ g)^n (b_0)$.

We define the function $h: A \to B$ as follows: \[h(a) = \begin{cases} g^{-1}(a), &\text{if } f(a) \text{ is the descendent of a lonely point,} \\ f(a) &\text{otherwise.} \end{cases}\] Note that $f(a)$ cannot be lonely itself. If $f(a)$ is the descendent of a lonely point, then $f(a) = f \circ g (b)$ for some $b$; since $g$ is injective, the element $g^{-1}(a)$ is well defined. Thus our function $h$ is well defined. We claim that it is a bijection from $A$ to $B$.

We first prove that $h$ is surjective. Indeed, if $b \in B$ is the descendent of a lonely point, then $h(g(b)) = b$; and if $b$ is not the descendent of a lonely point, then $b$ is not lonely, so there is some $a \in A$ such that $f(a) = b$; by our definition, then, $h(a) = b$. Thus $h$ is surjective.

Next, we prove that $h$ is injective. We first note that for any $a \in A$, the point $h(a)$ is a descendent of a lonely point if and only if $f(a)$ is a descendent of a lonely point. Now suppose that we have two elements $a_1, a_2 \in A$ such that $h(a_1) = h(a_2)$. We consider two cases.

If $f(a_1)$ is the descendent of a lonely point, then so is $f(a_2)$. Then $g^{-1}(a_1) = h(a_1) = h(a_2) = g^{-1}(a_2)$. Since $g$ is a well defined function, it follows that $a_1 = a_2$.

On the other hand, if $f(a_1)$ is not a descendent of a lonely point, then neither is $f(a_2)$. It follows that $f(a_1) = h(a_1) = h(a_2) = f(a_2)$. Since $f$ is injective, $a_1 = a_2$.

Thus $h$ is injective. Since $h$ is surjective and injective, it is bijective, as desired. $\blacksquare$


Problems

The Schroeder-Bernstein Theorem can be used to solve many cardinal arithmetic problems. For example, one may wish to show $|S|=\kappa$ for some cardinal $\kappa$. One strategy is to find sets $A,B$ such that $|A|=|B|=\kappa$ with injections from $A$ to $S$ and $S$ to $B$, thus concluding that $|A|=|S|=|B|=\kappa$. More generally, the Schroeder-Bernstein Theorem shows that the relation $|A|\leq|B|$ between cardinals $|A|$ and $|B|$ defined by "there is an injection $f:A\rightarrow B$" is a partial-order on the class of cardinals.

Introductory

Problem 1

Show that $\mathbb{Q}$ is countable.

Problem 2

Let $\kappa$ satisfy $\kappa\cdot\kappa=\kappa$. Show that $\kappa^{\kappa}=2^{\kappa}$.

Intermediate

Problem 1

We say a set of reals $O\subseteq\mathbb{R}$ is open if for all $r\in O$, there is an open interval $I$ satisfying $r\in I\subseteq O$. Show that the following sets are equal in cardinality:

  • $\mathbb{R}$
  • $2^{\aleph_{0}}$
  • $\{O\subset\mathbb{R}\mid O\text{ is open}\}$
  • $\{f:\mathbb{R}\rightarrow\mathbb{R}\mid f\text{ is continuous}\}$

See Also

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