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My Mathematical Analysis III professor gave me this problem:

Let $f:(0,1) \rightarrow f((0,1))$ be a continuous function in the standard euclidean metric space $($$\Bbb R$,$d_2$$)$ and let $\liminf_{x\rightarrow0} f(x)<\limsup_{x\rightarrow0} f(x)$, then prove that for every L $\in$ $(\liminf_{x\rightarrow0} f(x),\limsup_{x\rightarrow0} f(x))$, exists a sequence $x_{n}$ in $(0,1)$ that converges to $0$ and such that $\lim_{n\rightarrow \infty} f(x_{n}) = L$

I truly don't know how to prove it, if someone could help me i would be grateful.

Ayrcast2
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    Consider the set $E$ of those numbers $\lambda$ such that there exists a sequence $x_n \to 0$ with the property that $f(x_n) \to \lambda$. Then you should prove that $\liminf_{x \to 0} f(x) = \inf E$ and $\limsup_{x \to 0} f(x) = \sup E$. – Siminore Oct 14 '13 at 17:09
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    This is an excellent exercise to get acquainted with liminf/limsup in a concrete situation. By asking for a solution here instead of scribbling inequalities and wondering which arguments can go through, you decide to miss this opportunity. – Did Oct 14 '13 at 17:28
  • There is a fantastic description of $\limsup$ and $\liminf$ in the accepted answer to this question from which I have learned a lot. There isn't an answer to your question, but it should help a lot along the way – Keeran Brabazon Oct 14 '13 at 20:36
  • @Did: didn't you think that maybe i tryed to find an answer on my own for hours before posting here? – Ayrcast2 Oct 15 '13 at 13:16
  • You did? Right, then you will easily add to the question your thoughts and your failed tries about it (a practice which, on this site, is recommended). – Did Oct 15 '13 at 13:19
  • @Did: i thought to define A(L) = {x $\in$ (0,1) : f(x) = L} and to took a sequence x(n) = sup (B(0,1/n)$\cap$A(L)), the problem here is that i don't know how to prove that A(L) is not empty. – Ayrcast2 Oct 15 '13 at 13:42
  • Good starting point. Now, can you show that there exists y and z in B(0,1/n) such that f(y)<L<f(z)? – Did Oct 15 '13 at 13:46
  • @Did: i can tell that if L $\in$ [a,b] for every a,b such that liminf f < a < b < limsup f because f is continuous (intermediate value theorem), but how can i prove it in the open space? – Ayrcast2 Oct 15 '13 at 14:02
  • Reread the question: one assumes that liminf<L<limsup hence one is always in your L∈[a,b] case. – Did Oct 15 '13 at 14:04
  • @Did: lol, you are right, i was worrying about nothing! Thanks a lot. – Ayrcast2 Oct 15 '13 at 14:11
  • You might want to write down a solution of the question and to post it here. After a while, you may even accept it. That way, you can check your solution is allright and the site benefits from it. – Did Oct 15 '13 at 14:14

1 Answers1

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Given a continuous map $f \colon (0,1) \to \mathbf R$, put $m = \liminf_{x\to 0} f(x)$ and $M = \limsup_{x\to 0} f(x)$, and suppose that $-\infty \leqslant m < M \leqslant \infty$. Let $m < c < M$. Then there is a sequence $\{x_n\} \subset (0,1)$ such that $x_n \to 0$ and $f(x_n) \to c$.

Proof sketch. By definition of $m$ and $M$, and since $m < c < M$, there are sequences $\{y_k\},\{z_k\} \subset (0,1)$ such that $y_k, z_k \to 0$ and $m < f(y_k) < c < f(z_k) < M$ for every $k$.

Choose subsequences $\{y_{k_n}\}_n\subset\{y_k\}$ and $\{z_{k_n}\}_n\subset \{z_k\}$ such that $y_{k_n} < z_{k_n}$ for every $n$, which is possible since $\lim_{k\to\infty} y_k = \lim_{k\to\infty} z_k = 0$.

Since $f$ is continuous, the intermediate value theorem guarantees the existence of points $x_n$ satisfying

  • $y_{k_n} < x_n < z_{k_n}$ for every $n$, and

  • $f(x_n) = c$ for every $n$.

Thus, $\{x_n\}\subset (0,1)$ is the desired sequence. $\square$

Alex Ortiz
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