THEOREM PROOF Ax1. If p is a level then p is not above p Ax2. If each of P and Q
ID: 3119877 • Letter: T
Question
THEOREM PROOF
Ax1. If p is a level then p is not above p
Ax2. If each of P and Q is a level and P is above Q, then there exists a level X such that P is above X and X is above Q
Ax3. if P and Q are levels then P is above Q or Q is above P
Ax4. If P, Q, and R are levels, and P is above Q and Q is above R, then P is above R
Ax5. If P is a level, then there exists levels x and y such that y is above P and P is above x.
Ax6. If S1 and S2 are level sets such that (1) if x is a level the x is in S1 or x is in S2 and (2) if x is a level in S1 and y is a level in S2 then y is above x, then S1 has a top level or S2 has a bottom level
Definition 6: A level set M is called an interval provided there exist two levels A and B such that B is above A and x belongs to M if and only if x is in the segment (A, B) orx is A or x is B. In this case M is called the interval AB and is denoted [A, B].
Denition 7. The statement that a set M is nite means there is a positive integer n such that M contains exactly n elements.
Prove the following theorem
Theorem 3. If M is a nite level set then M does not have an accumulation level.
Explanation / Answer
Suppose M is the finite set;
that means;
for some n, total number of levels in M is n;
=> there exists a X for which level of X is above all other levels.
and there exists a Y for which level of every other levels in M is above Y.
So M has limit between X and Y.
that means M does not have any accumulation level.
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