Question

How would you prove this?
Given an arbitrary set E, a point z of E is said to be isolated if z is not a limit point of z. Prove that the set of all isolated points of E i countable.

Answer 1

maybe this helps to get ideas....
set is said to be countable if is finite or equivalent to N. (equivalent means one-to-one correspondance can be established)

Answer 2

Lecture Notes Exercise 12: Show that Q, the set of real rational numbers,
does not have the least upper-bound property.
Solution: To show this, we must show that there exists a set E ⊂ Q such
that E is nonempty and bounded above, but for which sup E does not
exist in Q. One example of such a set E is {x ∈ Q : x2 < 2}. This is
clearly nonempty and bounded above (by any member of Q greater than √2), and yet this set has no least upper bound, hence no supremum. For
consider the set of all positive rationals x such that x2 > 2. This is
precisely the set of upper bounds of E in Q (note well, that, crucially, the
real number x such that x2 = 2 does not belong to Q; indeed, we could
be doing exactly this proof using any irrational number in place of √2).
I claim that this set (call it F ) has no least element: for every p ∈ F, we
can find another q ∈ F such that q < p.
To do this, associate with every positive rational number p the number
q = p −
p2 − 2
p + 2
=
2p + 2
p + 2
Note first that all such b are positive. Note also that subtracting √2 from
both sides gives
q −
√2 =
(p − √2)(2 − √2)
p + 2
So pick a p ∈ F . Then the associated q is positive, but because p2 − 2 is
also positive, we know that q < p. At the same time, knowing that p−√2
is positive implies that q −√2 is too, and so we have shown that q ∈ F, as
we wanted to. Thus, we have shown that the set of upper bounds for E
has no least element in Q, which is the same as saying that sup E does not
exist in Q, implying that Q does not have the least-upper-bound property.
2. Is the set of real irrational numbers countable?
Solution: No. This follows immediately from the facts (noted in class
and shown in the lecture notes) that the reals are uncountable and the
rationals are countable. Since the reals are the union (indeed, finite union)
of the rationals and irrationals, the irrationals cannot be countable; if they
were, we would have to conclude that the reals were countable (being the
finite union of two countable sets), and they are not.
3. For x ∈ R1 and y ∈ R1, define
(a) d1(x, y) = (x − y)2
(b) d2(x, y) = |x − 2y|
(c) d3(x, y) = |x−y| 1+|x−y|
1
Determine for each of these whether it is a metric or not.
Solution:
(a) d1(x, y) is not a metric; it fails the triangle inequality. Consider
x = 2, y = 1, z = 0; then (x − y)2 +(y − z)2 = 2, while (x − z)2 = 4.
(b) d2(x, y) is also not a metric, for d2(1, 0) = 1 6= 2 = d2(0, 1).
(c) d3(x, y) is a metric. It is helpful to note that |x − y| is itself a metric;
it is therefore nonzero and positive for distinct x and y, zero for x = y,
and |x − y| = |y − x|, which implies that d3(x, y) is itself nonzero and
positive for distinct x and y, zero for x = y, and d3(x, y) = d3(y, x).
It remains to check the triangle inequality.
We want to show that
|x − y|
1 + |x − y|
+ |y − z|
1 + |y − z| ≥ |x − z|
1 + |x − z|
Let |x − y| = a, |y − z| = b, |x − z| = c. Thenwewant to show
a
1 + a
+
b
1+ b −
c
1 + c ≥ 0
a(1 + b)(1 + c) + b(1 + a)(1 + c) − c(1 + a)(1 + b)
(1 + a)(1 + b)(1 + c) ≥ 0
Note that the denominator is positive because a, b, and c are all
distances, and thus positive. So we just want to show that the
numerator is positive. It reduces to a + b − c+2ab + abc. The last
two terms are clearly positive; moreover, because a, b, and c satisfy
the triangle inequality we know that
a + b − c = |x − y| + |y − z| − |x − z| ≥ 0
and so we know that the numerator is positive. This shows that
d3(x, y) satisfies the triangle inequality; it is a metric.

Answer 3

but hete i snothing about isolated points

Answer 4

@Rohangrr