Question

Prove or disprove: If \(A\) is uncountable, then \(|A|=|\mathbb{R}|\)

Short answer

Disproved. There exist uncountable sets, such as the power set of the real numbers, that have a larger cardinality than the set of real numbers.

Step-by-step solution

Understanding Notation and Key Concepts

Before proving or disproving that if a set \(A\) is uncountable, then its cardinality equals the cardinality of the set of real numbers, it's crucial to understand some key concepts. The symbol \(|\cdot|\) represents the size or cardinality of a set. Thus, \(|A|\) refers to the size or cardinality of set \(A\), and \(|\mathbb{R}|\) refers to the size or cardinality of the set of real numbers.

Construct an Example

In order to disprove a statement, it's enough to find a counter-example. Consider the set of all sequences of natural numbers. This set is denoted as \(\mathbb{N}^\mathbb{N}\). This set is uncountable because there are infinitely many sequences, each of which can be infinitely long. Thus, this set is larger than any set of natural numbers.

Compare Cardinalities

It was Georg Cantor who showed that \(|\mathbb{N}^\mathbb{N}| = |\mathbb{R}|\), and \(|\mathbb{N}^\mathbb{N}| > |\mathbb{N}|\). It's important to note, however, that there is another uncountable set with a greater cardinality, namely the set of all subsets of \( \mathbb{R} \), also known as the power set of \( \mathbb{R} \), is strictly larger than \(|\mathbb{R}|\). Cantor's theorem states that for any set \(S\), the power set of \(S\) (i.e., the set of all subsets of \(S\)) has a strictly larger cardinality than \(S\) itself. Therefore, \(|\mathcal{P}(\mathbb{R})| > |\mathbb{R}|\).

Conclusion

Given the evidence, it's possible to disprove the statement. There are uncountable sets such as the power set of the real numbers whose cardinality is strictly larger than the cardinality of the set of real numbers. Therefore, not every uncountable set \(A\) has a cardinality equal to \(|\mathbb{R}|\).