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Dec 23, 2017 Researching Ioncube Php Encoder 8 Crack.

Aug 09, 2017 Welcome to my website, in this particular occasion We’ll show you regarding ioncube php encoder 8 crack. And after this, this is the 1st impression:Q:

Do we need to use axiom of choice in this proof about sets with infinite cardinalities?

How to prove that $\aleph_0$ is the least infinite cardinal number?

It is straight forward to see that $\aleph_0$ is the smallest ordinal number which is not finite. It is also a well-known fact that the smallest cardinal number which is not finite is $\aleph_0$.
But I have another proof to prove $\aleph_0$ as the least infinite cardinal number:

Since $\aleph_0$ is defined as the least cardinal number which is not finite, it follows from this definition that $\aleph_0$ is the least infinite cardinal number.

Now, my problem is when I want to prove that $\aleph_0$ is the smallest infinite cardinal number, is it right that I may use the axiom of choice to do it? If so, should I write another proof for that to prove that $\aleph_0$ is the smallest infinite cardinal number without the axiom of choice?

A:

First of all, you’re not allowed to use the axiom of choice in general when working with infinite cardinalities. But if you have an explicit definition of $\aleph_0$ that you like to work with (which is not what you do when you write proofs about the cardinality of $\mathbb{R}$ or $\mathbb{C}$), then you could write a proof without using the axiom of choice. For example, the following proof can be done without the axiom of choice using Kuratowski’s definition of the $\aleph$ numbers as the least cardinal number with $\omega$ many subsets.
Let $\kappa$ be an infinite cardinal number. Then $\kappa$ is the least infinite cardinal number if and only if $\kappa$ is the least cardinal number which is greater than $\omega$. The greatest infinite cardinal number is $\aleph_0$, since the cardinal $\aleph_0$ is greater than every cardinal. Now suppose that $\kappa$ is an infinite

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