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. .. Wrong, wrong, you’re wrong,
When you say all this
After all, I still barely live,
And I didn’t quite fall in love.
I myself, not that I don’t believe
But life has become not sweet to me,
I’m on my own, I can’t stand it anymore
And I’m already going crazy.
I’m crazy about the one next to me
I’m going crazy from the one that is far away
I’m going crazy, from those who are nearby,
I’m crazy about the ones I love.
I go crazy when everything is in moderation,
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Download Monster Hunter Frontier Offline Pc Downloadinstmank |Dansa.com / For the brave souls that want to play without needing an internet connection. can play offline.Q:
Why do we use the axiom of choice for finding a non-empty subset of a given set?
I’m having a hard time understanding why the axiom of choice is necessary in the following question,
Question: Is there a non-empty subset $X$ of the set of real numbers with the property that $\forall n \in \mathbb{N}, nx \in X$?
By non-empty I mean that it is $
ot\subseteq\emptyset$, and by $n \in \mathbb{N}$ that the $n$ is a natural number.
I’m thinking that if we can have a one-to-one correspondence between $X$ and the natural numbers, then $\mathbb{N} \to X$ would be injective and hence we’d have a proof without the axiom of choice.
My question is: what is the precise reason why we use the axiom of choice for this question?
A:
The axiom of choice is needed because you want to choose an element from every set that you look into (what is usually denoted by the symbol $\bigcup_{n\in\Bbb N}A_n$), not just every element of every set.
In your proposed line of argumentation, the point is that the natural number $0$ corresponds to the empty subset of $\Bbb R$, but the point is that there are larger and larger subsets with the property that $n\cdot x\in X$. This is not possible in a naive fashion, since there is no largest non-empty subset of $\Bbb R$ with this property.
That is, the property is a very strong property that actually requires a different proof than for every $n\in\Bbb N$ there is an $x\in X$ with $n\cdot x\in X$ (which in turn cannot be shown without the axiom of choice).
In fact, you can already see this in the statement of the problem: the statement only looks for a non-empty
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