Theorem: if $f$ is continuous on closed interval $I$ then it is also bounded. The proof is given by contradiction, assuming that for any $n$ it is possible to find $x_n$ such that it is greater than n $f(x_n)>n$. It is proven to be impossible by noting that because $x_n$ is bounded sequence it has a subsequence $x_{n_k}$ that is convergent and the limit again lies within the interval $I$. Why is it so? The second question is about the contradiction where it is stated that because the subsequence is convergent it is also bounded which is in contradiction to a fact that $|f(x_{n_k})|>n_k>k \\ \forall k \in N$. I don't understand why $n_k \ge k$?
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$I$ is presumably bounded as well as closed, so it's a compact interval. Every sequence in a compact set has a convergent subsequence, with limit in that compact set. For your second question, $n_k \ge k$ just obviously, since $n_k$ is a subsequence (this is a simple combinatorial exercise), but also you said something incoherently and irrelevant; what is irrelevant is that the sequence is bounded, all that matters is that $x_{n_k} \to x$, so $f(x_{n_k}) \to f(x)$ is a contradiction, since $f(x) < \infty$. – mathworker21 Aug 20 '19 at 14:50
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@GreyFox obv false – mathworker21 Aug 20 '19 at 23:47
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@mathworker21 Sorry, I wrote the proposition in a wrong way. The proposition is: if the sequence $(x_n)$ converges to $x$ and $x<a$ then $x_n<a$ for $n$ greater then some $n_0$. – Grey Fox Aug 21 '19 at 10:42
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Since the interval $I$ is closed, when a sequence $(x_n)_{n\in\mathbb N}$ of elements of $I$ converges, its limit belongs to $I$.
And whenever we have a subsequence $(x_{n_k})_{k\in\mathbb N}$ of a sequence $(x_n)_{n\in\mathbb N}$, the map $k\mapsto n_k$ is strictly increasing and therefore $n_1\geqslant1$, $n_2>n_1\implies n_2\geqslant 2$, $n_3>n_2\implies n_3\geqslant3$, and so on…
José Carlos Santos
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