Prove that if $A\subseteq B$, and $A$ is uncountable, then $B$ is uncountable

"Prove that if $A\subseteq B$, and $A$ is uncountable, then $B$ is uncountable"

I think that for an answer we could reason with cardinality as follows Suppose that $B$ is countable, then $\mid B\mid\leq\mid\mathbb{N\mid} $.

But if $A\subseteq B$, then $|A|\leq|B|\leq|\mathbb{N\mid}$ that is untrue if $A$ is uncountable, since any uncountable set should have higher cardinality then natural numbers.

Is it correct?

Solution 1:

$A\subseteq B$

Then the inclusion map $I :A \to B$ defined by $I(a) =a $ for all $a\in A $ is an one-to-one map.

Hence, $card(A) \le card(B) $

Given, $A$ is uncountable i.e $\aleph_{0} < card(A) $

Hence, $\aleph_{0} < card(A)\le card(B) $

So, $\aleph_{0} < card(B) $ implies $B$ is also uncountable.

Note: $\aleph_{0} = card (\mathbb{N})$