Prove that the function $f$ pointwise limit of a function sequence $(f_n)$ is continous, where $(f_n)$ follows a certain property.

Question

Let $f_n: I \to \Bbb{R}$ be a sequence of functions such that for all $(x_n)$ a sequence that conveges, the sequence $(f_n(x_n))$ converges. ($I$ is an interval)

$(f_n)$ converges pointwise to a function $f$ (take a constant sequence). I want to prove that $f$ is continous.

Any tips?

Honestly the only thing I could prove is that if $x_n$ converges to $x$, then $(f_n(x_n))$ converges to $f(x)$ I'm still very far from the result — Alex, Jan 08 '22 at 16:28
You could delete it. I've voted to close as duplicate, but the duplicate target gives a much clearer statement of the problem. — hardmath, Jan 09 '22 at 04:20

score -2 · Answer 1 · answered Jan 08 '22 at 16:22

-2

There is a rule that if interval $I$ is closed and finite, then $f$ is continuous and monotonous, with terms $\min$ and $\sup$

answered Jan 08 '22 at 16:22

HAD HAN

21

Prove that the function $f$ pointwise limit of a function sequence $(f_n)$ is continous, where $(f_n)$ follows a certain property.

1 Answers1