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Let A be an uncountable set, and let T be any non-empty set.Prove that $A \times T$ is uncountable

I firstly assume that $A \times T$ is countable, and show that it is uncountable by contradiction.

So, I started as,

Assume A is an uncountable set and T is any non-empty set, and there is some injective map $f:A \times T \to \mathbb{N}$.Addition to those we know that there is no such injective map $h:A \to \mathbb{N}$.

but after that in order to show that $A \times T$ is uncountable, either I have to consider the all possible cases one-by-one, or I have to show that the map $f$, somehow, depends on the map $h$, so since there is no $h$, there can not be $f$, but I'm totally stuck on that, because I can define $f$ depending on $h$, but I don't have to.

So I'm particularly look for proofs stems from this base that I gave, but, of course, if you have another kind of method to proof this statement, I would like to learn.

Note: I'm a freshman year Mathematics student, so if you give and explain the proof at the level of a first year Math student, I would appreciate that.

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    Have you proven every subset of a countable set is at most countable. Ax {1}~A is an uncountable subset. – 2017-02-10
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    @fleablood No, I did not, but I didn't understand what you implied either. – 2017-02-10
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    Basically, if f:AxT->N then restricting f to all (a,t) for a single constant value of t, creates a bijection from A to a subset of N. "Surely" a subset of N is countable. – 2017-02-10
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    "If f is injective from A into N then A is countable" is the same theorem as "every subset of a countable set is countable". Because the bijection of the superset, when restricted to the subset, is an injection of the subset into N. – 2017-02-10

2 Answers 2

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It's sometimes easier to use the theorem the if $B $ is countable and $C\subset B $, then $C $ is at most countable.

Alternatively this means if $C\subset B $ and $C$ is uncountable, then $B $ is uncountable.

I'll prove those at the end of this post.

Now let $t\in T$. Let $f:A\rightarrow A\times \{t\} $ via $f (a)=(a,t) $. $f $ is a bijection, so $A\times {t} $ is uncountable. And $A\times\{t\}\subset A\times T $. So $A\times T $ is uncountable.

So now we just need to prove that if $B $ is countable and $C\subset B $ then $C $ is at most countable.

Let $f$ be a bijection from N to $B $ and for indexing let $f (i)=a_i \in B $.

Define $g $ so that if $a_0\in C$ then $g (0)=a_0$. Otherwise try to see if $a_1$ in $C $.

More formally let $k_0 =$ the minimal $i $ so that $a_i \in C $. Let $g (0)=a_i $.

For $j >0$ let $k_j $ be the minimal $i $ so that $i >k_{j-1} $ and $a_i \in C $. Let $g (j)=a_i $.

(Basically we are just reindexing $C $) If $C $ is infinite this is a bijection from $C $ to the natural. So $C $ is countable. Otherwise the process ends and $C$ is finite (this is a bijection to a finite set of naturals)..

So that's it!

If $C \subset B $ and $B $ is countable, then $C$ is at most countable.

If $C \subset B$ and $C $ is uncountable, then $B$ being countable is impossible. So $B $ is uncountable.

$A\times\{t\} $ is uncountable.

And $A\times \{t\}\subset A\times T$.

So $A\times T$ is uncountable.

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Suppose $A \times T$ is countable. First, you need to assume that $f : A \times T \to \mathbb N$ is a bijection (this is what it means to be countable). Now pick some element $t \in T$. Doesn't the subset $A \times \{t\}$ have the same cardinality as $A$? So $A \times \{t\}$ is uncountable. Restricting your map $f$ to the subset $A \times \{t\}$ gives an injection $A \times \{t\} \to \mathbb N$, which contradicts the fact that $A \times \{t\}$ is uncountable. Hence $A \times T$ is uncountable.

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    Ok, first if $f$ is just injective, it is enough for us, I forgot to write in the question, thanks for reminding.Second, if you can prove that $A \times {t}$ is uncountable, it is already says that $A \times T$ is uncountable, and you did not give any proof for this. – 2017-02-10
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    Ax{t} = {(a,t)|a in A and t is t}. Let h ((a,t))= a. Then h is a bijection between A and Ax {t}. So Ax {t} is the same cardinality as A... which is uncountable. – 2017-02-10