How do we show if $|A|\le|B|$ then $|B|\not\lt|A|$ using Cantor-Schröder-Bernstein Theorem?

Question

The books states I have to make use of the Cantor-Berstein Theorem which states if $|A|\le|B|$ and $|B|\le|A|$ then $|A|=|B|$.

Attempt: Suppose $|A|\le|B|$, then $|A|\lt |B|$ or $|A|=|B|$. If this is true, then $|A| \not\lt|B|$ and $A\neq B$ is false. Hence $(|A|<|B|\lor A=B) \lor (|A|\not\lt|B|\land A\neq B)$ is true. Hence $(|A|<|B|\lor |A|\not\lt|B|\lor|A|\neq|B|)\land(|A|<|B|\lor|A|=|B|\lor|A|\neq|B|)$. Hence $(|A|\neq |B|)\land(|A|<|B|)$. Hence $|A|<|B|$. Hence $\neg(|B|<|A|)$. Hence $|B|\not\lt|A|$.

However, I did not use Cantor-Bernstein Theorem? Is my proof correct? What should I have done instead?

Answer 1

If $|B|<|A|$, then $|B|\le|A|$ and $|B|\neq|A|$ (according to textbook).

Assume $|A|\le|B|$, then by Cantor Bernstein Theorem if $|B|\le |A|$ and $|A|\le|B|$ then $|B|=|A|$; however, $|B|\neq|A|$. This is a contradition. Hence $|A|\not\le|B|$. Hence if $|B|\lt|A|$ then $|A|\not\le|B|$. Therefore by contraposition, if $|A|\le|B|$ then $|B|\not\lt|A|$.

Answer 2

If $|B|<|A|$, then there is an injection from $B$ to $A$. You’re assuming that $|A|\le|B|$, so there is also an injection from $A$ to $B$. The Cantor-Schröder-Bernstein theorem then says that $|A|=|B|$, contradicting the assumption that $|B|<|A|$. Thus, $|B|\not<|A|$. That’s all there is to it.

Answer 3

fleablood’s comment hits the nail on the head: you can’t just jump from $|A| < |B|$ to $\neg (|B| < |A|)$ because this assumes Cantor-Bernstein. There is no inherent contradiction between $|A| < |B|$ and $|B| < |A|$ that follows trivially from their definitions; the OP is subtly assuming there is a contradiction because the notation is suggestive (this makes it a good notation, but proof by notation alone is not a proof).

If you expand out the definitions, it just means that

1. there is an injection from $A$ to $B$,
2. there is an injection from $B$ to $A$,
3. there is no bijection between $A$ and $B$.

Cantor-Bernstein is precisely equivalent to saying that these three conditions cannot simultaneously occur. Without it, it’s entirely plausible that $|A| < |B|$ and $|B| < |A|$ at the same time.