# Showing a subset is uncountable [closed]

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

-

## closed as off-topic by Andres Caicedo, user127096, Claude Leibovici, tetori, Sami Ben RomdhaneApr 5 '14 at 8:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Andres Caicedo, user127096, Claude Leibovici, tetori, Sami Ben Romdhane
If this question can be reworded to fit the rules in the help center, please edit the question.

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

• 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.

-
• 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 $B$.
-

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.

-