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Let $ f : \mathbb{R} \rightarrow \mathbb{R} $ be a function which takes a convergent sequence and gives us a convergent sequence.

Show that $f$ is continuous.

So I saw a proof for this but I don't get it.

proof

We show that for all convergent sequences $a_n$ with limit $a$ it holds that $ \lim_{ n \rightarrow \infty } f(a_n) = f(a)$. Let $a_n$ be convergent with limit $a$. define $ b_n :=a_{\frac{n}{2}} $ if $n$ even and $ b_n := a $ if $n$ odd.

Now the unclear part

The following first equation is unclear:

And the last part with "it follows that" is unclear:

$ \lim_{ n \rightarrow \infty } f(b_n) = \lim_{ n \rightarrow \infty } f(b_{2n+1} ) = \lim_{ n \rightarrow \infty } f(a) = f(a) $. It follow that $ \lim_{ n \rightarrow \infty } f(a_n) = a. $

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  • $\begingroup$ Please also identify the source in which you found the proof you still need to share with us. $\endgroup$ – jordan_glen Jan 19 at 19:24
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The sequence $\bigl(f(b_{2n+1})\bigr)_{n\in\mathbb N}$ is a subsequence of the sequence $\bigl(f(b_n)\bigr)_{n\in\mathbb N}$ and therefore, since the limit $\lim_{n\in\mathbb N}f(b_n)$ exists, you have$$\lim_{n\in\mathbb N}f(b_n)=\lim_{n\in\mathbb N}f(b_{2n+1}).$$So, the limit $\lim_{n\in\mathbb N}f(b_{2n})$ also exists, but $b_{2n}=a_n$ and therefore $\lim_{n\to\infty}f(a_n)$ exists (and it is equal to $f(a)$).

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This proof centers on the Theorem:

If $ \{s_n\}\to s$ is a convergent sequence then for any sub sequence $\{s_{k_i}\}$, $\{s_{k_i}\}$ is a convergent sequence that converges to $s$.

So in this case you have:

$\{a_k\} \to a$ (given)

$\{a_k\} = \{b_2k\} \subset \{b_n\}$ (by the way we defined $\{b_n\}$.

$b_{2k+1} = a$ (by definition)

$\{b_{2k+1}\} \subset \{b_n\}$.

It's easy to show $\{b_n\} \to a$ [1]. And we know $\{a_k\} = \{b_{2k}\} \to a$. And $\{a\} = \{b_{2k+1}\} \to a$.

....

Now we are told that since $\{a_k\}=\{b_{2k}\}, \{a\}=\{b_{2k+1}\}, \{b_n\}$ all converge then

$\{f(a_k)\}=\{f(b_{2k})\}, \{f(a)\} = \{f(b_{2k+1})\}, $ and $\{f(b_n)\}$ all converge.

But $\{f(a_k)\} = \{f(b_{2k})\}\subset \{f(b_n)\}$, and $\{f(a)\} = \{f(b_{2k+1})\}\subset \{f(b_n)\}$, so they must all converge to the same value.

So $\lim_{k\to \infty} f(a_k) = $

$\lim_{k\to \infty}f(b_{2k}) =$ (because $a_k = b_{2k}$)

$\lim_{n\to \infty}f(b_n) =$ (because $\{f(b_{2k})\}\subset \{f(b_n)\}$)

$\lim_{k\to \infty}f(b_{2k+1})=$ (because $\{f(b_{2k+1})\}\subset \{f(b_n)\}$)

$\lim_{k\to \infty}f(a)$ (because $b_{2k+1} =a$)

$= f(a)$ (because $\{f(a)\}$ is a constant sequence)

And that's that.

For any $\{a_k\} \to a$ we have proven that $\{f(a_k)\} \to f(a)$ and that is the definition of continuous.

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[1]: If $a_n \to a$ and $b_{2k}= a_k$ and $b_{2k+1} = a$ then $b_n \to a$.

Pf: Let $\epsilon > 0$. There is an $N$ so that if $n > N$ then $|a_n - a| < \epsilon$. So there is an $M = 2N$ where if $n > M$ then

if $n$ is even then $\frac n2 > N$ and $|b_n - a|=|a_{\frac n2} - a| < \epsilon$

or if $n$ is odd then $|b_n - a| = |a-a| = 0 < \epsilon$.

So $b_n \to a$.

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Well the first equality of the first equation holds because the limit of a sequence (assuming it exists) is equal to the limit of any of its subsequence.

As for the "it follows that" part, You apply the same argument with $b_{2n}$ instead of $b_{2n+1}$. (I believed you have made a typo in the last equality btw).

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