Let $A$ be an infinite set and B a countable set, show that $\vert A \cup B \vert = \vert A \vert$

Question

Can someone please verify if my proof is correct, and add to it if it isn't?

Since we're given that $B$ is countable, we know $B$ is either finite or has the same cardinality as, say, the set $\mathbb{Q}$. So $\vert B \vert = \aleph_0$.

Similarly, if $A$ is infinite, then we know the result $\aleph_0 \leq \vert A \vert$.

This implies that $ \vert B \vert \leq \vert A \vert$.

Then $\vert A \cup B \vert = \vert A \vert$.

Am I missing something here?

Answer 1

You are missing something, yes. You only wrote down the following relations:

• $|B|=\aleph_0$, a relation that is not even true! There is nothing in the question that prevents $B$ from being a finite set.
• $|B|\leq |A|$
• $\aleph_0\leq |A|$

There is nothing here for you to conclude that $|A\cup B|=|A|$. For that, you would have to prove that $|A|\leq |A\cup B|$ (this should be easy) and that $|A\cup B|\leq |A|$ (this one should prove a little harder, as you need a surjection from $A$ to $A\cup B$ to prove this inequality).

Answer 2

I am not at all expert in large cardinals, but as far as my understanding goes, once you have proved $|B| = \aleph_0$, you then have (assuming that $A$ and $B$ are disjoint) $|A\cup B| = |A| + |B| = |A| + \aleph_0$, and the arithmetic cardinals is such that adding $\aleph_0$ to any infinite cardinal does not change said cardinal.

That being said, this "arithmetic" property is no less than what you are asked to show here, and I think the point of the question is to make you prove that. As @5xum suggested in his comment, you can define an injection $A \to A\cup B$ together with a surjection $A\to A\cup B$ and use the Cantor-Bernstein theorem.

To get a feel for how to do that, I would suggest to try to understand how a bijection between $\mathbb{N}$ and $\mathbb{N}^2$ works. Also note that without loss of generality, you can assume $B = \mathbb{Q}$ ou $\mathbb{N}$. Indeed since $|\mathbb{Q}| = |\mathbb{N}| = \aleph_0$, you have bijections $B\to\mathbb{N}$ and $B\to\mathbb{Q}$. So if you get a bijection $A\to A\cup \mathbb{N}$, you can then use the fact that the composition of bijections is a bijection. Finally you can assume that $\mathbb{N}\subset A$ (more precisely, there is a subset of $A$ which is in bijection with $\mathbb{N}$). This should help you mimic the construction of the case $\mathbb{N}^2$