Let $f$ be a continuous real-valued function on a closed interval [a,b]. I am misunderstanding a part of the proof of that continuous functions on closed intervals are bounded

Proof: Assume that f is not bounded on [a,b], then to each $n \in \Bbb N$ there exists an $x_n \in [a,b]$ such that $|f(x_n)| > n$. By Bolzano Weierstrass some subsequence of $x_n$, say $x_{n_k}$ converges to some real number $x_o$ which belongs to [a,b]. Since $f$ is continuous at $x_o$ then $lim f(x_{n_k}) = f(x_o)$, but $lim |f(x_{n_k})| = + \infty$, which is a contradiction, thus $f$ is bounded.

I need clarification why $lim |f(x_{n_k})| = + \infty$ since the existence of an $n \in \Bbb N$ such that $|f(x_n)| > n$ does not guarantee an existence of an $N \in \Bbb N$ such that for any $M > 0$ and $k>N$, $|f(x_k)| > M $ (the definition of a limit converging to infinity)

  • $\begingroup$ What you write in your second parraph is not the IVT but rather Weierstrass' first theorem on continuous functions: any cont. function in a bounded (or finite, if you will) closed interval is bounded there. The second Weierstrass theorem then tells us such a function gets its minimal and maximal values in the interval. $\endgroup$ – DonAntonio Aug 4 '16 at 22:53
  • $\begingroup$ @DonAntonio edited you are correct. $\endgroup$ – IntegrateThis Aug 4 '16 at 22:55
  • $\begingroup$ The nut of the argument is: If $(a_n)$ is not bounded above, then a subsequence of $(a_n)$ tends to $+\infty$. Can you show that? $\endgroup$ – Matthew Leingang Aug 4 '16 at 22:56
  • $\begingroup$ @MatthewLeingang yes but A-Priori would you not have to choose this subsequence to have this property, which the author did not. $\endgroup$ – IntegrateThis Aug 4 '16 at 22:57
  • $\begingroup$ Then you explained already...but you haven't yet realized it: since $\;|f(x_n)|>n\;$ , the subsequence that you get from applying B-W diverges towards infinity. That's all. $\endgroup$ – DonAntonio Aug 4 '16 at 22:57

Observe that you actually have


Now, who knows what happens with the infinite sequence $\;\{x_n\}\subset[a,b]\;$ , but by B-W theorem it has a convergent subsequence $\;x_{n_k}\xrightarrow[k\to\infty]{}x_0\;$ , and $\;x_0\in[a,b]\;$ because of closedness and

finitiness. Again, here you have $\;|f(n_1)|>n_1\;,\;\;|f(n_2)|>n_2>n_1,...$ etc.

  • 1
    $\begingroup$ I understand my mistake now which was silly. I was not associating the "n" in $|f(x_n)| > n $ to be the same value. Now I understand. $\endgroup$ – IntegrateThis Aug 4 '16 at 23:03
  • $\begingroup$ Also, the quoted proof seems to skip the part the subsequence $(x_{n_k})$ is the one that is chosen to converge. $\endgroup$ – Matthew Leingang Aug 4 '16 at 23:11

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