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How do I show if $A \subseteq B$, and $A$ is uncountable then $B$ is uncountable?

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closed as off-topic by Andrés Caicedo, user127096, Claude Leibovici, tetori, Sami Ben Romdhane Apr 5 '14 at 8:12

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I think it's trivial enough that you can take for granted, but you can do a simple contradiction. –  Grid Apr 5 '14 at 1:40
As a worshipper of simplicity, I have, as always, tried to give the simplest possible answer. Did I at least succeed in making it simpler than the others? –  Michael Hardy Apr 5 '14 at 1:57

3 Answers 3

up vote 2 down vote accepted
  • Proof 1: Here's a logical proof: $B$ contains $A$. Hence, $B$ contains an uncountable set, i.e., it contains a set with cardinal number larger than that of $\mathbb{N}$. Hence, as the cardinal number of $B$ is greater than or equal to that of $A$, $B$ has cardinal number larger than that of $\mathbb{N}$.

  • Proof 2: Proof by contradiction: If $B$ is countable, then $B$ has cardinal number equal than that of $\mathbb{N}$. However, as the cardinal number of $B$ is greater than or equal to that of $A$, and $A$ has cardinal number larger than that of $\mathbb{N}$, we have a contradiction.

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  • Proof 1: If $B$ is countable, there exists a surjection $f : \omega \rightarrow B$. If $A$ is empty, then it certainly is countable. So assume $A$ is not empty. Let $a \in A$. Define $f' : \omega \rightarrow A$ by

$f'(x) = \begin{cases} f(x) & \quad f(x) \in A \\ a & \quad \text{otherwise} \end{cases}$

$f'$ is a surjection. $A$ is countable.

  • Proof 2: (AC) As $A$ is uncountable, there is an injection $f : \omega_1 \rightarrow A$. Hence $f$ is an injection of $\omega_1$ into $B$.
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Suppose $B$ is countable, so $$ B=\{b_1,b_2,b_3,b_4,\ldots\}. $$ Now delete the ones that are not members of $A$. For example, you might have these ones left: $$ \{b_5,b_{13},b_{72},b_{986},b_{2003},\ldots\ldots\}. $$ Then let $$ \begin{align} a_1 & = b_5 \\ a_2 & = b_{13} \\ a_3 & = b_{72} \\ a_4 & = b_{986} \\ a_5 & = b_{2003} \\ & {}\,\, \vdots \end{align} $$ Then $A=\{a_1,a_2,a_3,\ldots\}$, so $A$ is countable.

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