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Theorem: If a finite set contains a real number, then it is a set of numbers Real.

Proof: By induction on the size of the set. Clearly true for sets of Size one. Suppose that the theorem is true for sets with n elements. Let's fix a set $A$ with $n + 1$ elements, $A=${$a_1,. . . , a_{n+1}$} and assume that $a_1$ is a real number. Then, the sets $A'=$ {$a_1,. . . , a_n$} and $A''= ${$a_1,. . . , a_{n-1}, a_{n+1}$} have n elements, and Contain some real number. Then $a_1,. . . , a_n, a_{n + 1}$ must be real numbers.

I think what is wrong is that we choose exactly a_1 as the real number, knowing that we can assume that there is a real number but we do not know what it is. Another thing is the fact of removing $a_{n + 1}$ and $a_n$ without knowing if this can be the real number that we have initially.

Hanul Jeon
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Nash
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    You seem to be assuming what you are trying to prove when you say "and assume that $a_1$ is a real number". How do you know that there is any real number in $A$? You don't! – MPW Dec 19 '16 at 22:04
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    This is essentially the same as this question: being a real number in your problem corresponds to having the single color that all horses have in the other question, and the argument fails for exactly the same reason. I’m going to mark this a duplicate even though the question is slightly disguised by changing horses to arbitrary objects and having some fixed color to being a real number. – Brian M. Scott Dec 19 '16 at 22:04
  • @MPW: No, that part is fine: the theorem is that if a set contains a real number, then it is a set of real numbers, and that is certainly true if it is a one-element set. – Brian M. Scott Dec 19 '16 at 22:05
  • @BrianM.Scott I do not understand that both problems are related, in case you mention failure for n = 1, but my problem remains for this case. – Nash Dec 19 '16 at 22:46
  • @Nash: The base case is fine for both arguments; it’s the induction step that fails for $n=1$. The reasoning in the induction step is fine if $n>1$, but it fails if $n=1$, because in that case $A$ does not contain two different elements besides $a_1$. The induction step works only if $A$ has at least $3$ elements, so that $A'$ and $A''$ have the real number $a_1$ in common and between them include every element of $A\setminus{a_1}$. \ The reason that we can assume that $a_1$ is real is that we’re assuming that $A$ contains at least one real number (since otherwise the theorem says ... – Brian M. Scott Dec 19 '16 at 22:53
  • ... nothing about $A$), and we can simply choose to apply the label $a_1$ to a real number in $A$. – Brian M. Scott Dec 19 '16 at 22:54
  • @BrianM.Scott Thak you very much. – Nash Dec 19 '16 at 23:11
  • @Nash: You’re very welcome. – Brian M. Scott Dec 19 '16 at 23:12
  • @BrianM.Scott Excuse me, I do not know English, that's why I find it harder to understand all this, for example I do not understand why the inductive hypothesis works only when A has at least 3 elements... – Nash Dec 19 '16 at 23:15
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    @Nash: Let me explain it for the specific case $n=2$. Then $A={a_1,a_2,a_3}$, and we assume that $a_1$ is real. The induction hypothesis is that if $B$ is a two-element set that contains at least one real number, then $B$ is a set of real numbers. Let $A'={a_1,a_2}$ and $A''={a_1,a_3}$; $A'$ and $A''$ are two-element sets, and each of them contains the real number $a_1$. The induction hypothesis therefore says that $A'$ and $A''$ are sets of real numbers. This means that $a_2$ is a real number (because it is in $A'$), and $a_3$ is a real number (because it is in $A''$). Thus, every ... – Brian M. Scott Dec 19 '16 at 23:20
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    ... member of $A$ is a real number, which is what we wanted to prove. If $n=3$, and $A={a_1,a_2,a_3,a_4}$ with $a_1$ real, we can do the same thing. The induction hypothesis is that every $3$-element set containing a real number is a set of real numbers. $A'={a_1,a_2,a_3}$ and $A''={a_1,a_2,a_4}$ are $3$-elements sets, and they contain the real number $a_1$, so by the induction hypothesis all of their members are real. For $A'$ this means that $a_2$ and $a_3$ are real, and from $A''$ it means that $a_2$ and $a_4$ are real, so every element of $A$ is real. But if $n=1$, so that ... – Brian M. Scott Dec 19 '16 at 23:23
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    ... $A={a_1,a_2}$, the only one-element subset of $A$ that is known to contain a real number is ${a_1}$. There is no subset of $A$ that we can use to conclude that $a_2$ is real. – Brian M. Scott Dec 19 '16 at 23:25
  • @BrianM.Scott I think this explanation is excellent, I do not know how to thank you for the patience and the fact of sharing your knowledge ... Thank you very much indeed and forgive so much insistence. Can I keep asking you about other issues? – Nash Dec 19 '16 at 23:29
  • @Nash: If they’re related to this question, you can certainly ask here. If they have to do with a different problem, it would be better to start a new question. – Brian M. Scott Dec 19 '16 at 23:30
  • @BrianM.Scott Okay, already this problem was clear to me! – Nash Dec 19 '16 at 23:33

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