1. Determine whether each of these sets is countable or uncountable. For those that are countable, exhibit a one-to-one correspondence between the set of natural numbers and that set.
a) integers not divisible by 3
d) the real numbers with decimal representations of all 1s or 9s

2. Show that if A and B are sets with the same cardinality, then the power set of A and the power set of B have the same cardinality.

3. Show that the set of real numbers that are solutions of quadratic equations ax^2 + bx + c = 0, where a,b, and c are integers, is countable.

4. Let U be an uncountable universal set. Prove or disprove the following statements:
a) If A is a subset of U is uncountable, the complement of A is countable.
b) If A is a subset of U is countable, the complement of A is uncountable.

5. Let f: X -> Y be 1-1. Show that if Y is countable, then X is countable.

Thanks A Million
Anu

2. Do you understand the distinction between countable and uncountable?
Every subset of a countable set is countable.
Every superset of an uncountable set is uncountable.
The set of integers, Z, is countable.
The set of reals, R, is uncountable.

Given those facts, answer #1 & #4.

If $f:A \mapsto B$ define the image function $\vec f:P\left( A \right) \mapsto P\left( B \right),\;\left( {C \subseteq A} \right)\left[ {\vec f(C) = \left\{ {b \in B:\exists a \in A \wedge f(a) = b} \right\}} \right]$.
Now show that if $f:A \mapsto B$ is bijective then $\vec f:P\left( A \right) \mapsto P\left( B \right)$ is bijective.
That proves #2.

For #3 note that $Z \times Z \times Z$ is countable and every quadratic equation has at most two real roots.

3. is this a joke proof??

5) $f:X\rightarrow Y$ is 1-1.
then x corresponds to a unique element of y..
but since Y is countable, this implies that X must be countable..

haha..

4. Originally Posted by anu
...
4. Let U be an uncountable universal set. Prove or disprove the following statements:
a) If A is a subset of U is uncountable, the complement of A is countable.
b) If A is a subset of U is countable, the complement of A is uncountable.
..
Can we consider consider the set of Complex Numbers as our universal set?
if so, then that disproves a).
say, if we take the subset irrational numbers.. still, the complement includes bi where b is a real number..

for b) if we take a countable subset from an uncountable set, then the complement of the countable set must be uncountable..
say this:
Let $A \subset U$ be countable; and suppose $A^c$ is countable..
clearly, A and $A^c$ are disjoint, so that
$A \cup A^c = U$ and the union of countable sets are countable which is a contradiction to the assumption that U is the universal uncountable set.. Hence, $A^c$ must be uncountable..

5. Originally Posted by Plato
If $f:A \mapsto B$ define the image function $\vec f:P\left( A \right) \mapsto P\left( B \right),\;\left( {C \subseteq A} \right)\left[ {\vec f(C) = \left\{ {b \in B:\exists a \in A \wedge f(a) = b} \right\}} \right]$.
Now show that if $f:A \mapsto B$ is bijective then $\vec f:P\left( A \right) \mapsto P\left( B \right)$ is bijective.
That proves #2.
what? can u explain what you wrote please?