There is a big difference between 1 and {1}. 1 is an element of {1} but not the other way around. Also, the empty set is an element of 1 but not of {1}.I dont know how to write the symbols thats why I wrote 0 and 1.
Let's agree to write 0 for the empty set . All other sets can be written using curly braces and commas. You can also write "x in y" for .
By definition, nothing is an element of the empty set. You keep repeating this mistake even after it was pointed out.
It seems that you don't understand the set notation. First, x is not the same as {x}. Also, I am not sure you understand the difference between an element and a subset.So they are the same. Since 0 (emptyset) is the "only" element that both have. Correct ? For 0 and 2 I talk for the first universe.
Well I officially quit. I dont get it noway. If 0 and {0} -zero stands for emptyset- are not equal in the first universe, 0 and {0,{0}} are also not equal both in the first and in the second universe e.t.c. I cant figure out how to prove the type.
Thanks a lot for your time. I think you ' re right. I will fail logic. It is too difficult for me.
Thanks one more. You're very kind.
Sorry just one last question because I think I got it.
In the second universe {1} and 2 dont have the same elements but they are equal correct ??
empty set and one are IN {1}
empty set and one are IN 2
but their elements are different... ???
You are welcome.
Your misunderstanding so far may have come from something very basic, like the notation used to describe sets. A wrong understanding of this may render everything else incomprehensible. If this is the case, then your problems are understandable, and they do not mean that you have problems with logic in general.
No, {1}={{0}} and 2={0,{0}}={0,1} are non-equal sets because they are written in different ways.In the second universe {1} and 2 dont have the same elements but they are equal correct ??
empty set and one are IN {1}
empty set and one are IN 2
Also, {1} and 2 have different elements. I wrote a table before that for each of the three elements of the universe in b) says whether this element belongs to {1} and to 2. In particular, the empty set is not an element of {{1}}: it is not listed explicitly. But the empty set is an element of 2.
I would recommend making sure you understand the notation used to describe sets as well as the concepts of an element and of a subset. You can read about them in a textbook or in Wikipedia. Nobody can move forward until these things are absolutely clear.
Sorry {1} doesnt include (0,1)? It includes only number one? not zero?
From Wikipedia: (http://en.wikipedia.org/wiki/Natural_number)
2 = {1} = {{{ }}}, etc.
I'm sure now I got it. Maybe I dont express myself in a propre way but I think that's the solution.
Sorry got stuck on how to prove by induction the above type for every finite set.
Ok for n=1 lets take universe A={a}. So:
case first : P={emptyset}
case second: P={(a,a)}
Lets say that its valid for n=k.
We ll prove for n=k+1 & we have universe:A={a0,a1,a2,...,a(k),a(k+1)}.
How do I go on?
Thanks in advance for any help.
Writing "case first" does not constitute a proof. When I grade(d) a homework, I am willing to allow sketchy proofs for someone who has consistently demonstrated that they understand the material and are more than able to fill in missing details. In this situation, however, the case P={emptyset} (which probably is supposed to mean P=emptyset) is impossible, which is not clear at all from your version.Ok for n=1 lets take universe A={a}. So:
case first : P={emptyset}
case second: P={(a,a)}
One can find the minimum of k+1 elements by finding the minimum m of the first k elements and then comparing m with the (k+1)st element.We ll prove for n=k+1 & we have universe:A={a0,a1,a2,...,a(k),a(k+1)}.
How do I go on?
Proving ∃y∀xP(y,x) under assumption
(∀x(P(x,x)∧∀x∀y∀z[(P(x,y)∧P(y,z))→P(x,z)]∧∀x∀y[P(x,y)∨P(y,x)]) is equivalent to finding a minimum of the universe with respect to the reflexive, transitive and total relation P. I wrote that the problem of finding a minimum of k+1 elements can be reduced to finding a minimum of k elements and of 2 elements.