Difference between revisions of "2006 Romanian NMO Problems/Grade 10/Problem 1"
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==Solution== | ==Solution== | ||
− | I claim that <math>f(A)=|A|</math>, for all <math>A\in \mathcal P(M)</math>. Clearly this function works; we must now show that it is the only function with the two given properties. We shall do this by | + | I claim that <math>f(A)=|A|</math>, for all <math>A\in \mathcal P(M)</math>. Clearly this function works; we must now show that it is the only function with the two given properties. We shall do this by proving that any such function <math>f</math> that satisfies both properties must be <math>f(A)=|A|</math>. |
− | + | We have <math>f(\emptyset) = f(\emptyset) + f(\emptyset)</math>, hence <math>f(\emptyset) = 0</math>. Also, if <math>A \subset B \subset M</math>, then <math>f(B) = f(A) + f(B-A)</math> since <math>B = B \cup A</math>, <math>A = B \cap A</math>, and <math>B-A = B\Delta A</math>. | |
− | + | Let <cmath>A_{0} = \emptyset \subsetneq A_{1} \subsetneq \dotsb \subsetneq A_{n-1} \subsetneq A_{n} = M</cmath> be an increasing sequence of subsets of <math>M</math>, such that <math>A_{i} = A_{i-1} \cup \{a_{i}\}</math> for all <math>i=1,\dotsc,n</math>. Then <math>|A_{i}| = i</math> for all <math>i</math>. | |
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− | {{ | + | Note that <math>f(a_{i}) \ge 1</math> for any <math>i</math>, from property (a). Hence <cmath>f(A_{0}) = 0 < f(A_{1}) < \dotsb < f(A_{n-1}) < f(A_{n}) \le n \;,</cmath> which implies <math>f(A_{i}) = i</math> for all <math>i</math>. |
==See Also== | ==See Also== |
Revision as of 18:15, 7 May 2012
Problem
Let be a set composed of
elements and let
be its power set. Find all functions
that have the properties
(a) , for
;
(b) , for all
, where
.
Solution
I claim that , for all
. Clearly this function works; we must now show that it is the only function with the two given properties. We shall do this by proving that any such function
that satisfies both properties must be
.
We have , hence
. Also, if
, then
since
,
, and
.
Let be an increasing sequence of subsets of
, such that
for all
. Then
for all
.
Note that for any
, from property (a). Hence
which implies
for all
.