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.


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. $

  • $\begingroup$ 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

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 $f(a)$).


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.


[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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.