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just need a bit of help understanding this answer.

Let $(x_i)$ be a sequence of distinct elements in a metric space, and suppose $x_i \rightarrow x$. Let f be a 1-2-1 map of the set of $x_i$'s into itself. Prove that $f(x_i) \rightarrow x$

so intuitively the result theyre asking to prove makes sense. their answer is

"working via contradiction, $\exists$ a sequence $(x_{i_{k}})~\text{and}~\epsilon > 0~s.t: d(f(x_{i_{k}}),x) \geq \epsilon.$ since $x_i \rightarrow x$ and $\{x_i: i\in\mathbb{N}\}$ is a sequence of distinct elements,then $\{i\in\mathbb{N}: d(x_i,x) \geq \epsilon\}$ is finite. Hence $\{f(x_{i_k}): k = 1,2,... \}$ is finite which contradicts the fact that f is 1-2-1"

So my understanding of this solution goes as this.

we have a set in which there's a sequence of points which converge to a limit point x. we apply a injective mapping to this set which basically move's points around, for sake of ease up and down a linear line. we're assuming that the mapped points dont converge to x. which means they converge else where or just plain ole diverge.

So the set of points that are not in x's neighbourhood is finite? but surely we could choose epsilon to be artbitarily large to encompass them too? or these points are just isolated points in which $n<M$ in the definition of convergence.

Either way, ok we have a finite set of points not in the neighbourhood of x which i can sort of understand since $f(x_{i_k})$ isn't converging to x. and i guess i understand that this makes $\{f(x_{i_k}): k = 1,2,... \}$ finite, but i dont understand how this draws a contradiction.

if someone could explain that would be great.

Thanks for taking the time to read.

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  • $\begingroup$ I’m supposing that “1-2-1” is to be read as one-to-one. ( ? ) $\endgroup$
    – Lubin
    Oct 15, 2018 at 2:12

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Since $f$ is one-to-one, since $f$ maps each element of the sequence $(x_n)_{n\in\mathbb N}$ into another element of that sequence, and since the inequality $d\bigl(f(x_{i_k}),x\bigr)\geqslant\varepsilon$ holds for each $k$, then the inequality $d(x_n,x)\geqslant\varepsilon$ holds infinitely many times. But this contradicts the assumption that $\lim_{n\to\infty}x_n=x$.

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