Schröder Bernstein Theorem: Proof
Consider the notation:
Define, for each , the order of it, denoted by , to be the number of such preimage(s) which exist. In a similer way, we’d be able to define the order of an element , i.e., by considering the sequence .
Now define, for each ,
Notice that if the order is infinite, is also infinite. Because, otherwise would have to have a finite order. On the other hand, if and is infinite, then exists and has an infinite order; call the latter one . This means, maps the infinite order elements of bijectively onto the infinite order elements of .
Next, if , then the order of is sheer . Similer to the above para, if for , the order is , as the order is non-zero, exists and it must have order . To formally show it you need a tedious inductive reasoning!
Last, if , then the order is , and so, exists and the order of is sheer (looking upon as an element of ). Conversely, similer to above, if there is an element of order in , take and the order of is indeed . All that you need to convince a sceptic is a long, tedious, involved induction!!
What we learn from what precedes is that the one-one function maps the infinite order elements onto infinite order elements, odd order onto odd order, and even order onto even order; a simple set theory reveals that is a one-one map from onto . This completes the proof.
|Title||Schröder Bernstein Theorem: Proof|
|Date of creation||2013-03-22 16:40:36|
|Last modified on||2013-03-22 16:40:36|
|Last modified by||sauravbhaumik (15615)|