Is $B - A$ always similar to $B$ if $A$ is countable and $B$ is uncountable?

[Dustinsfl]

If $A$ is a countable set and $B$ an uncountable set, prove that $B - A$ is similar to $B$.Case 1: $|A| = n\in\mathbb{Z}^+$
Since $B$ is uncountable, $|B| = 2^{\aleph_0}$.
Then $|B - A| = 2^{\aleph_0} - n = 2^{\aleph_0}$.
Therefore, $B - A$ is equinumerous to $B$, and hence $B - A$ is similar to $B$.Case 2: $|A| = \aleph_0$
Again, we have $|B - A| = 2^{\aleph_0} - \aleph_0 = 2^{\aleph_0}$
Therefore, $B - A$ is equinumerous to $B$, and hence $B - A$ is similar to $B$.

Does this work?

 

[Evgeny.Makarov]

What does "similar" mean?

Since $B$ is uncountable, $|B| = 2^{\aleph_0}$.

This is wrong. Also, even for finite sets, |B - A| is not necessarily |B| - |A|.

 

[Plato2]

If $A$ is a countable set and $B$ an uncountable set, prove that $B - A$ is similar to $B$.

There is completely trivial proof if A is a subset of B.
You know that the union of two countable sets is countable.
You also know that $B=A\cup(B-A)$. What if $B-A$ were countable?