If I prove that $|A| > |\mathbb{N}|$, does it mean that $A$ is uncountably infinite?

Question

My intuition is that if something is larger than $\mathbb{N}$ then it follows that it is larger than $\mathbb{Q}$ as there is a bijection from $\mathbb{N}$ to $\mathbb{Q}$. So if a set is larger than $\mathbb{Q}$, which is the boundary before going into $\mathbb{R}$, then it must be uncountably infinite.

But I am not sure if this is valid.

Answer 1

Yes, if $|A| > |\mathbb N|$, then $A$ is uncountable (and infinite). [This is not the Cointinuum Hypothesis. ]

On the other hand, you wrote $|A| > \mathbb N$. What does that mean? Do you mean $|A| > k$ for all $k \in \mathbb N$? In that case, it does not follow that $A$ is uncountable. For example $A = \mathbb N$ satisfies this.

Another warning. $|A| > |\mathbb N|$ means: there is an injective map $\mathbb N \to A$, but there is no injective map $A \to \mathbb N$. It is not enough to show there is a map $\mathbb N \to A$ that is injective but not surjective. That would only show $|A| \ge |\mathbb N|$

Answer 2

Your set is infinite, because there is an injection $\mathbb{N}\to A$; it is not countably infinite, because this would mean $|A|=|\mathbb{N}|$.

Mentioning $\mathbb{Q}$ and $\mathbb{R}$ is irrelevant.

Answer 3

If you have proven that its cardinality is greater than $\omega$, then you have indeed proven that the set $A$ is uncountable.

However, note that this does not mean the cardinality of $A$ is equal to the cardinality of $\mathbb{R}$. We do not know if there are any sets whose cardinality is between that of the integers and the continuum.