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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. $

The sequence $bigl(f(b_{2n+1})bigr)_{ninmathbb N}$ is a subsequence of the sequence $bigl(f(b_n)bigr)_{ninmathbb N}$ and therefore, since the limit $lim_{ninmathbb N}f(b_n)$ exists, you have$$lim_{ninmathbb N}f(b_n)=lim_{ninmathbb N}f(b_{2n+1}).$$So, the limit $lim_{ninmathbb N}f(b_{2n})$ also exists, but $b_{2n}=a_n$ and therefore $lim_{ntoinfty}f(a_n)$ exists (and it is equal to $f(a)$).

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_{kto infty} f(a_k) = $

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

$lim_{nto infty}f(b_n) =$ (because ${f(b_{2k})}subset {f(b_n)}$)

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

$lim_{kto infty}f(a)$ (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 $b_n to a$.

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).