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| ``Re: a post for entry "ordinal number"''
by djao on 2004-11-19 13:34:50 |
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| You're making this too difficult.
2) take x={0}. S has three elements: 0, {0}, and {0,{0}}, with the ordering relation x<y being "x is a proper subset of y". Recall the definition of proper subset: "x is a proper subset of y" means "(x is a subset of y) AND (x is not equal to y)."
0 is a subset of x because the empty set is always a subset of every set. 0 is a proper subset of x because 0 does not equal {0}.
{0} is a subset of x because {0} = x. However {0} is NOT a proper subset of x, because proper subset by definition means nonequal.
{0,{0}} is not a subset of x because {0,{0}} has two elements and x only has one element.
Therefore, out of the three elements of S, precisely one element of S is a proper subset of x. This one element is 0. The set consisting of all such elements is {0}, which does in fact equal x, as claimed.
> since in this entry one opeates with notation "<", then it is
> indirectly assumed that a<a can't be
This is correct. The definition of proper subset ensures that "x propersubset x" can't be. I don't see why you have a problem with proper subset being an example of <
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