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In the theorem of Schroder-Bernstein, it is assumed that, given two sets $A$ and $B$, and there is an injective map from $A$ to $B$ and an injective map from $B$ to $A$. It then concludes that there is bijection between $A$ to $B$.

My question here is based on hypothesis of the theorem, although it may not be closely related to the theorem, but it came to me when I first re-read the statement.

Question: Given non-empty sets $A$ and $B$, is it true that there is always an injective map from $A$ to $B$ or $B$ to $A$.

p Groups
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  • How do we prove this? any hint? – p Groups Jun 30 '16 at 03:39
  • Assuming the axiom of choice, this is always true. – Kyle Gannon Jun 30 '16 at 03:40
  • It's not possible for a function to be injective if it's domain has more elements than its range. – Jared Jun 30 '16 at 03:44
  • In that case, I would think this statement is true, either the number of elements of $A > B$ or $B > A$ or they are are equal. In which case it's always possible that $A\mapsto B$ can be injective or $B \mapsto A$ can be injective. I'm not sure how to handle continuous domains though. – Jared Jun 30 '16 at 03:48
  • See also the infinitely many linked questions here: http://math.stackexchange.com/questions/linked/268942 – Asaf Karagila Jul 04 '16 at 15:38

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In expansion of my comment, this is true assuming the axiom of choice. If $A$ and $B$ are sets, they are in bijection with cardinals $\kappa$, $\mu$ respectively. Then, $\kappa \in \mu$, $\mu \in \kappa$, or $\mu = \kappa$. In each case, have have an injection (i.e. inclusion\equality).

Kyle Gannon
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Your statement is:

$(*)$ Given non-empty sets $A$ and $B$, there is an injective map from $A$ to $B$ or from $B$ to $A$.

The statement $(*)$ is equivalent to the axiom of choice, as follows.

$AC \implies (*):$

Consider the set $P$ of all injective maps from a subset of $A$ to $B$. Partially order $P$ by set inclusion. (We're thinking of the functions that belong to $P$ as sets of ordered pairs, as usual.)

Using Zorn's lemma, $P$ has a maximal element $f.$ By maximality, either (1) the domain of $f$ is all of $A,$ in which case $f$ is an injective map from $A$ to $B,$ or (2) the range of $f$ is all of $B,$ in which case $f^{-1}$ is an injective map from $B$ to $A.$

$(*)\implies AC:$

Let $S$ be any set. We'll show that S can be well-ordered.

By Hartogs' theorem, there is an ordinal $\eta$ so large that there is no injective map from $\eta$ to $S$. By $(*)$, there must therefore exist an injective map $i$ from $S$ to $\eta.$ It follows that $i(x) \lt i(y)$ defines a well-ordering of $S$.

Mitchell Spector
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It is easily shown for Von-Neumann ordinals. But each set is in bijection with its cardinal (a Von-Neumann ordinal) so it is sufficient to show for Von-Neumann ordinals.

Jacob Wakem
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I will only answer in terms of discrete domains, where the number of elements is countable. It's very simple, if the domain contains fewer elements than the range then an injection is not possible. It's a simple pigeonhole principle problem where $n$ is the amount in the domain and $m$ is the amount in the range therefore any map which maps all elements of the domain to the range must include at least one element of the domain which maps to $>1$ elements of the range.

Other than that, I guess I don't understand your question. Certainly if the domain has fewer elements than the range then an injection is possible and certainly if the number of elements in the domain and range are the same then an injection is always possible (basically you can re-map all values of the domain to $\{1...N\}$ and all values of the range to $\{1...N\}$ then map $f(x) = x$).

So, in conclusion, I would think that yes, all possible 2-sets of domains would permit a mapping which is injective one way or or the other, either from $A\mapsto B$ or $B\mapsto A$ but what would be the point of proving such a theorem???

Jared
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