Show that a set S has infinite elements if and only if it has a
subset U such that (1) U does not equal to S and (2) U and S have
the same cardinality.
Let A be a subset of all Real Numbers. Prove that A is closed
and bounded (I.e. compact) if and only if every sequence of numbers
from A has a subsequence that converges to a point in A.
Given it is an if and only if I know we need to do a forward and
backwards proof. For the backwards proof I was thinking of
approaching it via contrapositive, but I am having a hard time
writing the proof in...
1. Let A be an inductive subset of R. Prove that {1} ∪ {x + 1 |
x ∈ A} is inductive.
2. (a) Let n ∈ N(Natural number) and suppose that k 2 < n
< (k + 1)2 for some k ∈ N. Prove that n does not have a square
root in N.
(b) Let c ∈ R \ {0}. Prove that if c has a square root in Z,
then c has a square root in...
Let A ⊆ C be infinite and denote by A' the set of all the limit
points of A.
Prove that if z ∈ A' then there is a non-trivial sequence of
elements in A that converges to z
Let X be a subset of R^n. Prove that the following are
equivalent:
1) X is open in R^n with the Euclidean metric
d(x,y) = sqrt((x1 - y1)^2+(x2 - y2)^2+...+(xn - yn)^2)
2) X is open in R^n with the taxicab metric
d(x,y)= |x1 - y1|+|x2 - y2|+...+|xn - yn|
3) X is open in R^n with the square metric
d(x,y)= max{|x1 - y1|,|x2 - y2|,...,|xn -y n|}
(This can be proved by showing the 1 implies 2 implies 3)...