I'm not sure what you mean by numerically equivalent. One way of defining two sets to have the same cardinality if and only if there exists a bijection between the two.For infinite sets \(\displaystyle A\) and \(\displaystyle B\) to be numerically equivalent. Is bijective function
\(\displaystyle f:A \rightarrow B\)
the only requirement that needs satisfying?
There is nothing new here. If \(\displaystyle \text{card }A=\alpha,\text{card }B=\beta\) then \(\displaystyle \alpha\leqslant\beta\Leftrightarrow \exists f:A\overset{\text{injection}}{\longrightarrow}B\)There is apparently another way of comparing set cardinalities, with cardinal numbers but I never learned anything on the topic and someone more knowledgeable on the matter will have to discuss it.
Are you thinking of \(\displaystyle A\) and \(\displaystyle B\) being countable?If \(\displaystyle \text{card }A=\alpha,\text{card }B=\beta\) then \(\displaystyle \alpha\leqslant\beta\Leftrightarrow \exists f:A\overset{\text{injection}}{\longrightarrow}B\)
Numerically equivalent set by definition is |A|=|B| where the sets are infinite. Please see more details in the question I put to Drexel.I'm not sure what you mean by numerically equivalent. One way of defining two sets to have the same cardinality if and only if there exists a bijection between the two.
There is apparently another way of comparing set cardinalities, with cardinal numbers but I never learned anything on the topic and someone more knowledgeable on the matter will have to discuss it.
(Note that when I say if and only if its merely a logical statement used in every definition, sometimes implicitly.)
Well, you have made it very clear for me to understand.If I recall correctly, the definition is that \(\displaystyle |A| = |B| \text{ if } \exists f:A \overset{\text{bijection}} {\longrightarrow} B\), and \(\displaystyle |A| \leq |B| \text{ if } \exists f:A \overset{\text{injection}} {\longrightarrow} B\), as drexel has pointed out.
(This is the definition for any A,B - not just countable sets.)
Yes, if \(\displaystyle f: A \rightarrow B\) surjective then this implies \(\displaystyle A \geq B\). However, the general technique is that if \(\displaystyle f_1:A \rightarrow B\) is an injection and \(\displaystyle f_2:B \rightarrow A\) is also an injection, then \(\displaystyle |A| \leq |B|\) and \(\displaystyle |B| \leq |A|\) so \(\displaystyle |A| = |B|\).Well, you have made it very clear for me to understand.
Drexel runs at hypersonic speed when I am still crawling. I only knew that if \(\displaystyle f \) is injective, the \(\displaystyle |A| \leq |B|\), but I couldn't see how that automatically translate to \(\displaystyle f\) being bijective.
Now, I think you mean if \(\displaystyle f\) is bijective, then \(\displaystyle f:A\rightarrow B \Rightarrow |A|\leq |B| \), and if \(\displaystyle f \) is surjective, then \(\displaystyle |A|\geq |B|\). Then only after putting those two together,
\(\displaystyle |A|\leq |B|\) and \(\displaystyle |A|\geq |B| \Rightarrow |A|=|B| \) .
Yah?
(Rofl)... you don't need to go the whole hog and find a bijection!