The sum of an uncountable number of positive numbers
For any M, there exists a finite sub-family B of A such that the sum of B is at least M. Proof: Assume that A +, the positive members of A, is uncountable (otherwise the theorem is obviously false). A + = ⋃nAn, where An = {a ∈ A | a ≥ 1 n}. Since the union of countably many finite sets is countable, one of the An must be infinite.
Normal + Connected -> Uncountable - Mathematics Stack Exchange
Normal + Connected -> Uncountable. Here's a solution (to the exercise "prove that if X is a space with more than one element, and is normal and connected then X is uncountable"): by Urysohn's lemma, given A and B, closed and disjoint in X, there exists a continuous function from X into [0, 1] such that f(A) = 0, f(B) = 1. If X were countable ...
Proving a set is uncountable - Mathematics Stack Exchange
10. If A were countable, then f((0, 1)), which is a subset of A, would also be also countable. Since f: (0, 1) → A is injective then f: (0, 1) → f((0, 1)) is a bijective function, so (0, 1) would also be countable. But (0, 1) is in fact uncountable, so A is also uncountable. Share.
analysis - Proving the open interval $(0,1)$ is uncountable ...
2. You can do this by showing that there is a bijection between (0, 1) (0, 1) and R R. Two sets are equivalent (have equal cardinalities) if and only if there exists a bijection between them. R R is uncountable. So by showing that there exists a bijection from (0, 1) (0, 1) to R R, you thereby show that (0, 1) (0, 1) is uncountable.
Uncountable vs Countable Infinity - Mathematics Stack Exchange
2015年11月5日 · My friend and I were discussing infinity and stuff about it and ran into some disagreements regarding countable and uncountable infinity. As far as I understand, the list of all natural numbers is countably infinite and the list of reals between 0 and 1 is uncountably infinite. Cantor's diagonal proof shows how even a theoretically complete ...
$\\mathbb{Q}_{p}$ is uncountable - Mathematics Stack Exchange
Dietrich Burde. 135k 8 92 163. Add a comment. 2. The set of expansions ∑∞ i=0aipi ∑ i = 0 ∞ a i p i, with 0 ≤ai <p 0 ≤ a i <p is uncountable. The proof is literally the same as the proof that the interval [0, 1] [0, 1] is uncountable. Have a look into …
The set of natural number functions is uncountable
I thought the set of natural number functions would be of the same cardinality as the countably infinite product of $\\mathbb{N}$, which is countable. Each natural number function can be identified...
Is every connected metric space with at least two points …
As the topic, how to prove that every connected metric space with at least two points uncountable? Of course i know the definition that a countable set mean there is a bijection between the set and the positive integer. Connected is opposite of disconnected where the set can partition into two disjoint open sets.
real analysis - $\mathbb{R}$ is uncountable, Abbott's proof ...
There is this proof in Abbott's book (Understanding Analysis) on page 25, that I am failing to understand. Theorem is that R is uncountable. And here is how the author proceeds to prove it (I know there is an easier proof by Cantor): To show that a set is uncountable, we need to show that there is no 1-1,onto function of the form: f: N → R.
Proving that R is uncountable - Mathematics Stack Exchange
2019年3月28日 · By construction, this sequence is different from all the others. This contradicts the assumption that this set of sequences is countable. Hence, $[0,1]$ must be uncountable, and therefore $\mathbb{R}$ must be uncountable.