Correcting a Previous Blog Post
by djmathman, Dec 8, 2018, 3:45 AM
On January 4th of this year, I posted a solution to the following problem from my Combo 3 notes back at AwesomeMath 2013.
Problem. Let
be a nonempty set, and let
be an increasing function on the set of subsets of
, meaning that
if
. Prove that there exists
, a subset of
, such that
.
Unfortunately, it turns out that my solution is dead wrong. In the proof that my chosen set satisfied
, I said
This sentence is quite bogus, as
is just the value of the function
at
, not the image
(which doesn't even typecheck!). The rest of the solution is rendered invalid as a result, and I actually believe the approach outlined in the solution is doomed to fail.
Here is, as far as I can tell, an actual solution.
Solution
Problem. Let








Unfortunately, it turns out that my solution is dead wrong. In the proof that my chosen set satisfied

Quote:
Let
, so that there exists
such that
.






![$f[T] = \{f(x):x\in T\}$](http://latex.artofproblemsolving.com/d/4/4/d44712db41e54c2dbed5a2e0e1f2e1fe381c3228.png)
Here is, as far as I can tell, an actual solution.
Solution
Denote by
the collection of subsets
such that
; note that
is nonempty since it contains
.
Claim:
is upwards closed with respect to
; in other words, if
, then
.
Proof.
implies
; now
, and so
. 
Consider the partially ordered set
of all elements in
under the partial order
. We will show that every chain in
has an upper bound in
. To prove this, suppose
is some chain in
. Let
Note that
for all
, so
is certainly larger (in the sense of the partial order) than all the other elements in
. It remains to show that
. This is not hard, and actually resembles the proof from above: for all
, we have
; unioning over all such
yields
and so
.
As a result,
satisfies the axioms of Zorn's Lemma, and so we know that
has a maximal element. Call this element
. Then
, but since
is maximal and
, this inclusion cannot be strict; as a result, the only possibility is
.





Claim:




Proof.





Consider the partially ordered set







![\[
M = \bigcup_{C\in\mathcal C} C.
\]](http://latex.artofproblemsolving.com/e/b/0/eb0a10ba76c8a61b257a4c6ec5c0b55de8356d37.png)








![\[
f(M)\supseteq \bigcup_{C\in\mathcal C} C= M,
\]](http://latex.artofproblemsolving.com/8/4/8/848d5b50fe897ebb3ebfd9fa529c9abe9b2df78a.png)

As a result,







This post has been edited 5 times. Last edited by djmathman, Dec 8, 2018, 3:49 AM