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Algebra/A question about proofs in college algebra

Question
The following is a proof of a false theorem, the required from me is to determine the flaw in the arguments

Theorem: Let n be a natural number. Any set of at most n elements numbers contains at most one element.

Corollary: There are no finite sets with more than one element.

Proof of Theorem: Suppose the theorem is false. Let n be the smallest natural number for which it is false. Clearly n doesn't equal 1 since the theorem is true for n = 1. Let S be a set with no more than n elements for which the assertion is false. Since n is the smallest natural number for which the assertion is false, it is true for numbers less than n. So the set S must have exactly n elements. Let the elements of S be x_1, x_2, x_3, ..., x_n . Let T_i be the set containing every element of S except for x_i ; so its elements are

x_1, x_2, x_3, x_4,...., x_{i-1}, x_{i+1},..., x_n
Since each T_i has less elements than S, the assertion must be true of T_i . But then we have

x_1=x_2=x_3=x_4=x_{i-1}=x_{i+1}=x_n

But since this is true for all choices of i, we have with a different choice, say j

x_1=x_2=x_3=x_4=x_{j-1}=x_{j+1}=x_n

Since each of these leaves out only one of the x_k , we see that the two sets of equalities together show that all the x_k are equal. So, all the x_k are all equal. But then the assertion is true of S, contradicting the choice of S.

The flaw in the arguements, when we say that all the elements of T_i are equal, we assume that i is an unique element and no element in S is equal to i, but this could be false.

Am I right? Or there is another flaw?

The flaw is in the hidden assumption that the least such n is 3 or greater. In fact , the least such n is n = 2

If you try to carry out the argument with only two elements in S , Ti and Tj won't have any elements in common to establish the equality. That's where it fails.

Algebra

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