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Problem

Assume that $x_{n}<x_{n+1}(n=1,2,\cdots)$ and $\lim\limits_{n \to \infty}x_n=x$. Prove $$x>x_n$$ holds for all $n=1,2,\cdots.$

Proof

$\forall k \in \mathbb{N+}$,$\exists n \in \mathbb{N_+}: k<n$.We obtain $$x_k<x_n.\tag1$$ Fix $k$ and let $n \to \infty$. We obtain $$x_k \leq \lim_{n \to \infty}x_n=x.\tag2$$ $(2)$ holds for all $k$, thus $x$ is a upper bound of $x_n$, namely $$x_n \leq x. \tag3$$ But we may prove $\forall n \in \mathbb{N_+}:x \neq x_n$. If $\exists k\in \mathbb{N_+}:x=x_k$, then $$x_{k+1}>x_k=x,$$ which contradicts $(3).$ Therefore $$x>x_n.$$

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mengdie1982
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1 Answers1

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Basically right, with a particular inaccuracy.

Your first sentence states that there is some particular $n$ with $k < n$; but then you immediately let $n \to \infty$ without ever proving that there are infinitely many such $n$. That doesn't matter for your proof, though; you might just as well have started with "Fix $k$. Then for all $n > k$, we have $x_k < x_n$ inductively." Once you've got that, you're fine to take the limit on the right-hand side (as long as that's something you know the proof of, perhaps from your lecture notes; if it's not, you definitely need to prove it).

A different method of proof would be as follows.

Suppose the contrary: that there is $k$ with $x \leq x_k$. Then $x_{k+1}$ has $x < x_{k+1}$. But then all $i$ with $k+1 < i$ have $|x_i - x| > (x_{k+1} - x)$ inductively, so the limit of $(x_i)_{i > k+1}$ cannot be $x$ since they are uniformly bounded away from $x$. The limit of any subsequence of a convergent sequence is the limit of the sequence; this is the contradiction.

Patrick Stevens
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