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In this answer, @Hyperplane defined an infinite set as follows:

Well a set $M$ is infinite if for every finite subset $U \subsetneq M$ there exists a $x \in M, x\notin U$. Consequently, you will be able to construct a sequence $(x_n)_{n \in \mathbb{N}} $ of distinct elements in $M$. Therefore $|M| \ge \aleph_0 $

Then @Andrés E. Caicedo commented here:

This is not true as written. Replace $\subsetneq$ with $\subseteq$, and it will be fine now.

I think that @Hyperplane's definition of an infinite set is wrong. My reasoning is:

For every subset $U \subsetneq M$, we always have $\exists x \in M, x \notin U$ by the definition of $\subsetneq$. So @Hyperplane just restated the property of $\subsetneq$ and said nothing about the definition of infinite set.

When I replace $\subsetneq$ with $\subseteq$ in @Hyperplane's answer as suggested by @Andrés E. Caicedo, I even find the new definition more awkward. My reasoning is:

For $U=M$, there does NOT exist any $x \in M, x\notin U$. Thus the statement Well a set $M$ is infinite if for every finite subset $U \subseteq M$ there exists a $x \in M, x\notin U$ is wrong. Let alone define an infinite set.

Could you please check whether my above reasonings are correct?

Akira
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  • It is awkward, yes. It would be clearer if expressed as "$M$ is infinite if for every finite $U$ with $U\subseteq M,$ we have $U\subsetneq M.$" – Cameron Buie Sep 10 '18 at 11:26
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    @Cameron: Or, simply put, a set is infinite, if it is not equal to any of its finite subsets. – Asaf Karagila Sep 10 '18 at 12:33
  • @CameronBuie I have asked a question at https://math.stackexchange.com/questions/2907637/define-a-countably-infinite-subset-of-x for several days, but have not received any answer. Although I have some questions that are not answered such as this one. I'm very in need of getting an answer for this question. Could you please help me check it out? – Akira Sep 11 '18 at 23:27

3 Answers3

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Your reasoning about the original failed definition is correct.

Your critique of the modified definition misses the point slightly. You seem to be missing the condition that $U$ has to be finite; so if $M$ is not finite you cannot choose $U=M$ as a finite subset of $M$.

On the other hand, the outcome of this is that the modified definition simply says that $M$ is infinite iff $M$ is not a finite subset of $M$. Or in other words:

A set $M$ is called infinite if and only if it is not finite.

If we already know what "finite" means, this is a good definition. But bringing non-proper subsets into it just muddies the waters.

(On the other hand, if we don't already have a definition of "finite", then neither of the proposed definitions make sense).

hmakholm left over Monica
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  • Hi @Henning, I have asked a question at https://math.stackexchange.com/questions/2907637/define-a-countably-infinite-subset-of-x for several days, but have not received any answer. Could you please help me check it out? – Akira Sep 11 '18 at 03:37
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You are correct to think that the first given definition is wrong and you have gavin a good reason for it. However, for the corrected definition, you are not. Indeed, in order to chose $U=M$, $M$ must be finite.

nicomezi
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  • Hi @nicomezi, I have asked a question at https://math.stackexchange.com/questions/2907637/define-a-countably-infinite-subset-of-x for several days, but have not received any answer. Could you please help me check it out? – Akira Sep 11 '18 at 08:14
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The point is that if you try to choose $U=M$ and find a candidate for $x$ you fail - there aren't any elements to fit the definition of $x$, and since there are such $x$ for any finite $U$ it follows that $U=M$ cannot be finite.

Mark Bennet
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  • Hi @Mark, I have asked a question at https://math.stackexchange.com/questions/2907637/define-a-countably-infinite-subset-of-x for several days, but have not received any answer. Could you please help me check it out? – Akira Sep 11 '18 at 06:04