(2) Cantors Paradox demonstrates that there is no set of all sets. We know that every set has a cardinality, and so its natural to ask if there is a set of all possible cardinalities. The answer to this is also no. Your task is to prove it. More precisely, Let X be a set whose elements are sets, and suppose X has the property that every set has the same cardinality as some element of X. In other words, suppose X is a set that contains a representative of every possible cardinality. Show that X cant exist. Hint: Consider the cardinalities of T = U S and P(T). SEX

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(2) Cantors Paradox demonstrates that there is no set of all sets. We know that every set
has a cardinality, and so its natural to ask if there is a set of all possible cardinalities.
The answer to this is also no. Your task is to prove it. More precisely,
Let X be a set whose elements are sets, and suppose X has the property that every
set has the same cardinality as some element of X. In other words, suppose X is
a set that contains a representative of every possible cardinality. Show that X cant
exist. Hint: Consider the cardinalities of T = U S and P(T).
SEX
(3) Consider the cubic equation x + bx2 + cx + d = 0. Suppose b, c, d are constructible.
Prove that if the equation has a rational root then the other two roots must be
constructible. Note this is a partial converse to 12.3.22
(You're not allowed to use a cubic formula like Cardano's etc to solve the equation.)
x2
(4) Consider the ellipse :
a²
= 1
62
and the line : cx + dy
: (s, t) be
= 0 and let P =
a point of intersection.
Transcribed Image Text:(2) Cantors Paradox demonstrates that there is no set of all sets. We know that every set has a cardinality, and so its natural to ask if there is a set of all possible cardinalities. The answer to this is also no. Your task is to prove it. More precisely, Let X be a set whose elements are sets, and suppose X has the property that every set has the same cardinality as some element of X. In other words, suppose X is a set that contains a representative of every possible cardinality. Show that X cant exist. Hint: Consider the cardinalities of T = U S and P(T). SEX (3) Consider the cubic equation x + bx2 + cx + d = 0. Suppose b, c, d are constructible. Prove that if the equation has a rational root then the other two roots must be constructible. Note this is a partial converse to 12.3.22 (You're not allowed to use a cubic formula like Cardano's etc to solve the equation.) x2 (4) Consider the ellipse : a² = 1 62 and the line : cx + dy : (s, t) be = 0 and let P = a point of intersection.
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