ProphetesAI is thinking...
denumerable
MindMap
Loading...
Sources
denumerable
denumerable, a. Math. (dɪˈnjuːmərəb(ə)l) [f. denumerate v. + -able. Cf. G. abzählbar, Fr. dénombrable.] Of a set: infinite but countable; capable of being put into a one-to-one correspondence with the set of finite integers or natural numbers; also more widely, either finite or countably infinite; e...
Oxford English Dictionary
prophetes.ai
Definition of denumerable (countable) set When we say that a set $S$ is denumerable, that is, there is a bijection $S \to \omega$, do we mean that there _exists_ such a bijection or do we mean that we have one and are...
Denumerable means there exists a bijection between the given set and the set $\mathbb {N}$.
prophetes.ai
What's the basic steps to show a set is denumerable? For example, $\mathbb{N}$ is denumerable.
A set $A$ is denumerable iff there exists (at least) one injective function $f:A\to\Bbb N$ and one injective function $g:\Bbb N\to A$.
prophetes.ai
If A is a denumerable set, and there exists a surjective function from A to B, then B is denumerable I am having some trouble solving the following homework question and some help would be greatly appreciated!! Q: Pr...
So, prove first that a set $S\ne \emptyset $ is denumerable if, and only if, there exists a surjection $f:\mathbb N \to S$. Now, if $g:A\to B$ is surjective and $A$ is denumerable, then there exists a surjection $f:\mathbb N \to A$.
prophetes.ai
Trying to show that set of all one-element subsets of a denumerable set is denumerable Let $A$ be a denumerable and put $X = \\{ B : B \subset A, \; \; |B| =1 \\} $. Then $X$ is denumerable: I know there is a bijecti...
On the other hand, if we take $A= \mathbb{Z}$ (the set of all integers), then $X$ is denumerable, but $g$ fails to surject onto $X$ (since, in particular
prophetes.ai
Prove or disprove: If $A$ and $B$ are denumerable, then $A - B$ is denumerable Prove or disprove: If $A$ and $B$ are denumerable, then $A - B$ is denumerable Can someone give me a hint as to how to prove/disprove thi...
As A-B is a subset of A, A-B is denumerable.
prophetes.ai
Infinite subset of Denumerable set is denumerable? > **Possible Duplicate:** > An infinite subset of a countable set is countable If $B$ is a denumerable set and $C$ is a subset of $B$ and is infinite, $C$ is denu...
Since $B$ is denumerable, there is a bijection $f: B \rightarrow \mathbb{N}$.
prophetes.ai
Denumerable and infinite sets > If $A$ is an infinite set and $B$ is denumerable, $A$ is equipotent with the union $A\cup B$. How to prove this?
If $B'$ is infinite, let $D$ be a denumerable subset of $A$. (This requires you to prove that the union of two disjoint denumerable sets is denumerable; I trust you know how to do that)
prophetes.ai
Prove or disprove: If $A \subseteq B$ and $B$ is denumerable, then $A$ is denumerable Claim: If $A \subseteq B$ and $B$ is denumerable, then $A$ is denumerable Proof: Assume $A \subseteq B$ and $B$ is denumerable, th...
Suppose $A$ is a subset of $B$ and $B$ is denumerable. So we can enumerate the elements of $B$ as a sequence $x(1), x(2), ..., x(n),.$. Take $A = \mathbb{N}$ and $B = \mathbb{R}$, then $A$ is a subset of $B$ and $A$ is denumerable while $B$ is not.
prophetes.ai
How to show if A is denumerable and $x\in A$ then $A-\{x\}$ is denumerable **My thoughts:** If $A$ is denumerable then it has a bijection with $\mathbb{N}$ So therefore $A\rightarrow \mathbb{N}$. Then x is a single o...
Suppose that $f:\mathbb{N}\to A$ is a bijection, where $f(n)=x$ for some $n\in\mathbb{N}$. Define $$ g(k)=\left\\{\begin{array}{} f(k)&\text{if }k\lt n\\\ f(k+1)&\text{if }k\ge n\\\ \end{array} \right. $$ Then $g:\mathbb{N}\to A-\\{x\\}$ is a bijection.
prophetes.ai
Example to disprove If $A\subseteq B$ and $B$ is denumerable, then $A$ is denumerable I want to see if my counterexample is valid: Let $A=\\{5,6,7\\}$ and $B= \mathbb{N}$ Then, $B$ is denumerable, but $A$ is not. ...
Yes, that is a counterexample!
prophetes.ai
Giving examples of denumerable sets My question reads: Give an example of denumerable sets A and B, neither of which is a subset of the other, such that (a) A ∩ B is denumerable (b) A-B is denumerable. I am not ...
Then their intersection is denumerable. Construction like this, where their intersection is a set well known to be denumerable, is the easiest way to go.
prophetes.ai
Prove or disprove that $\mathbb{Q}-\mathbb{Z}$ is denumerable My question states: prove or disprove that $\mathbb{Q}-\mathbb{Z}$ is denumerable.
$\mathbb Q$ is denumerable, and so is any subset of it. More generally, any subset of a denumerable set is denumerable.
prophetes.ai
Prove that a denumerable set can be partitioned into two denumerable subsets I was wondering if this "proof" is sufficient in demonstrating a that a denumerable set $A$ can be partitioned into two denumerable subsets ...
Your argument is circular: you’ve _started_ with denumerable sets that form a partition, when what you need to show is that such sets exist. Pick a partition of $\Bbb N$ into two denumerable sets; there’s one that’s especially obvious and easy to describe.
prophetes.ai