Cantor-Schröder-Bernstein Theorem

Solved Problems

Solution.

Since \(A\) is a subset of \(B,\) there exists an injection \(A \to B,\) so we have \(\left| A \right| \le \left| B \right|.\) Likewise, since \(B \subseteq C,\) we find that \(\left| B \right| \le \left| C \right|.\) As a result, we obtain the double inequality

By condition, \(\left| A \right| = \left| C \right|.\) Therefore, we have

Using the Cantor-Schröder-Bernstein theorem, we get

Solution.

Since \(\left| \mathbb{R} \right| = \left| {\left( {0,1} \right)} \right|,\) we compare the cardinalities of \({{\mathbb{N}^{\mathbb{N}}}}\) and \({\left( {0,1} \right)}.\)

To build an injection \({\mathbb{N}^{\mathbb{N}}} \to \left( {0,1} \right),\) consider an arbitrary infinite sequence \(\left( {{a_1},{a_2},{a_3}, \ldots } \right) \in {\mathbb{N}^{\mathbb{N}}}\) where \({a_i} \in \mathbb{N}.\) This sequence is mapped to a number in the open unit interval:

Clearly, this mapping is injective.

To define \(\left| {\left( {0,1} \right)} \right| = \left| \mathbb{R} \right|,\) we use the fact that every real number has a unique infinite decimal representation. In this case, a finite decimal is converted into infinite one like

Let a real number \(b \in \left( {0,1} \right)\) be represented as

We can map the number \(b\) to the following infinite sequence:

This mapping is also injective.

Consequently, by the Cantor-Schröder-Bernstein theorem,

Solution.

Since \(\left| \mathbb{R} \right| = \left| {\mathcal{P}\left( \mathbb{N} \right)} \right| = {2^{\left| \mathbb{N} \right|}},\) we have

Recall (see Countable and Uncountable Sets) that

Hence,

Solution.

According to Heine definition, a real function \(f\left( x \right)\) is said to be continuous at a point \(x \in \mathbb{R}\) if for any sequence \(\left\{ {{q_n}} \right\}\) such that

it holds that

It is also known that for every \(x \in \mathbb{R},\) there exists a sequence of rational numbers \({q_n}\) that converges to \(x.\) Therefore, any continuous real-valued function \(f\) is determined by its values on the set of rational numbers \(\mathbb{Q}.\)

We denote the set of all continuous functions \(f:\mathbb{R} \to \mathbb{R}\) by \(C.\)

By restricting \(f\) to the rationals, we get a function \(f^{\prime}:\mathbb{Q} \to \mathbb{R}\) such that \(f\left( q \right) = f^{\prime}\left( q \right)\) for any \(q \in \mathbb{Q}.\) The mapping \(f \to f^{\prime}\) is injective, so

Since \(\left| {{\mathbb{R}^{\mathbb{Q}}}} \right| = \left| {{\mathbb{R}^{\mathbb{N}}}} \right| = \left|\mathbb{R}\right|\) (see Example \(3\)), we have

From the other side, there also exists an injection \(\mathbb{R} \to C.\) For example, we can map any real number \(\alpha \in \mathbb{R}\) to the constant function \({f_\alpha }: \mathbb{R} \to \mathbb{R}\) given by

All functions \(f_\alpha\) are continuous. Then, by the Cantor-Schröder-Bernstein theorem,