ONT Re: Set Theory
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Note 11
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| Elementary Set Theory
|
| Existence of Sets
|
| This section is concerned with the existence of sets and
| with the initial steps in the construction of functions
| and other relations from the primitives of set theory.
|
| III. Axiom of Subsets.
|
| If x is a set
| there is a set y
| such that for each z,
| if z c x, then z in y.
|
| 33. Theorem. If x is a set and z c x, then z is a set.
|
| Proof. According to the axiom of subsets, if x is a set
| there is y such that, if z c x, then z in y, and
| hence by definition 1, z is a set. (Observe that
| this proof does not use the full strength of the
| axiom of subsets since the argument does not
| require that y be a set.) þ
|
| 34. Theorem. 0 = |^| $U$
|
| and $U$ = |_| $U$.
|
| Proof. If x in |^| $U$, then x is a set and, since 0 c x, it follows from 33
|
| that 0 is a set. Then 0 in $U$ and each member of |^| $U$ belongs to 0.
|
| It follows that |^| $U$ has no members. Clearly (that is, by theorem 26),
|
| |_| $U$ c $U$. If x in $U$, then x is a set, and by the axiom of subsets
|
| there is a set y such that, if z c x, then z in y. In particular, x in y,
|
| and since y in $U$ it follows that x in |_| $U$. Consequently $U$ c |_| $U$
|
| and equality follows. þ
|
| 35. Theorem. If x =/= 0, then |^| x is a set.
|
| Proof. If x =/= 0, then for some y, y in x. But y is a set,
|
| and since |^| x c y by 32, it follows from 33 that
|
| |^| x is a set. þ
|
| JLK, Gen Top, pages 256-257.
|
| John L. Kelley, 'General Topology',
| Van Nostrand Reinhold, New York, NY, 1955.
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