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This is actually a generalized version I wrote of a homework question that intrigued me:

Let $f$ be continuous in $[0,1]$ and $\forall x\in [0,1], \ f(x)=f(1-x)$. If $r\in [0,1]$ and $1-r$ are the only sub limits of $a_{n}$ then $f(a_{n})$ converges.

(the original question states $r=\frac{1}{3}$)

I solved the original question using a method shown in class and tried to implement it to prove the generalized version (see my own answer). I was wondering if there's a better way.

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  • $\begingroup$ Again: please do not use a half sentence as your title. $\endgroup$ – Arturo Magidin Nov 13 '10 at 22:08
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    $\begingroup$ @Arturo: I'm not a native English speaker and I have no idea how to write the title differently. I can't write the entire question because it's too long so I wrote the important part. I wrote it the same way you edited my previous question, feel free to edit it again... $\endgroup$ – daniel.jackson Nov 14 '10 at 7:34
  • $\begingroup$ I'm not a native speaker either. I don't want to come in and rewrite your posts, which is why I only added ellipses in the previous one. But perhaps something like "Convergence of $f(a_n)$ when $(a_n)$ has only two limit points" or the like. The general gist, but not the particulars, would be the idea. $\endgroup$ – Arturo Magidin Nov 14 '10 at 23:00
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We claim that $f(a_n) \to \xi = f(r) = f(1-r)$. If $r = \tfrac{1}{2}$ then $(a_n)$ must be convergent (why?). Since $f$ is continuous $f(a_n)$ converges too.

So assume that $r \neq \tfrac{1}{2}$. WLOG we can assume that $r \lt \tfrac{1}{2}$. Consider the subsequence $(a_{n_k})$ consisting of the elements of $(a_n)$ that are less than $\tfrac{1}{2}$ and the subsequence $(a_{n_l})$ consisting of the elements of $(a_n)$ that are greater than or equal to $\tfrac{1}{2}$ and convince yourself that they are in fact subsequences and that they exhaust the $a_n$. In particular $a_{n_k} \to r$ and $a_{n_l} \to 1-r$. Now all that's left to do is to observe that since $f$ is continuous, $f(a_{n_k})$ and $f(a_{n_l})$ are eventually in any open interval around $\xi$. Since they exhaust $(a_n)$ we must have $f(a_n) \to \xi$.

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Here's my own attempt at solving this (comments are appreciated!):

We denote $b_{n}=f(a_{n})$ and by Bolzano-Weistrauss, $b_{n_{k}}\to b$. The subsequence $a_{n_{k}}$ might not converge, but again by Bolzano-Weistrauss there exists $a_{n_{k_{m}}}$ that converges to either $r$ or $1-r$ (because those are the only sublimits of $a_{n}$).

  1. if $a_{n_{k_{m}}}\to r$ then $b_{n_{k_{m}}}=f(a_{n_{k_{m}}})\to f(r)$ (because $f$ is continuous in $[0,1]$).
  2. Similarly, if $a_{n_{k_{m}}}\to 1-r$ then $b_{n_{k_{m}}}=f(a_{n_{k_{m}}})\to f(1-r)$.

But since $\forall x\in [0,1], \ f(x)=f(1-x)$, we get $f(r)=f(1-r)$.

To finish the argument we note that $b_{n_{k_{m}}}$ is a subsequence of $b_{n_{k}}\to b$, thus $b_{n_{k_{m}}}\to b$ and we conclude that $b=f(r)=f(1-r)$.

Since $b$ was chosen arbitrarily we get that $b_{n}$ has a single sublimit which implies $b_{n}\to f(r)$

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It is quite obvious that if $r$ and $1-r$ are the only sublimits of $\{a_n\}$ in the interval $[0,l]$, then for any $\delta \gt 0$, $(r-\delta, r+\delta)\cup (1-r-\delta,1-r+\delta)$ contains all but finitely many terms of the sequence. Let $\epsilon>0$. Choose $\delta>0$ such that $|f(x)-f(r)|\lt \epsilon$ for all $x\in (r-\delta, r+\delta)$. Then by symmetry of $f$, also $|f(x)-f(r)|\lt \epsilon$ for all $x\in (1-r-\delta, 1-r+\delta)$. It follows that $$|f(a_n)-f(r)|\lt \epsilon$$ for all but finitely many $n$'s. This proves that $\lim_{n\to\infty} f(a_n)=f(r)$.

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