Exists a continuous function $f: \mathbb R \to \mathbb R$ with $f(x_n) = y_n, \, \forall n$ and $f(x) = y$

Question

Let $x_n \to x$ and $y_n \to y$ in $\mathbb R$ such that $x_n \neq x_m, \, \forall n \neq m$.

How can I show the existance of a continuous function $f: \mathbb R \to \mathbb R$ with $f(x_n) = y_n, \, \forall n$ and $f(x) = y$?

I've trying to solving this problem in the last few days, however I didn't have any idea. I'm looking for a hint to solve this question.

Help?

Answer 1

Define $f$ by linear interpolation. Without loss of generality, we may assume that $\{x_n\}$ in increasing. On the interval $[x_n,x_{n+1}]$ define $f$ as $$ f(x)=y_n+\frac{x-x_n}{x_{n+1}-x_n}\,(y_{n+1}-y_n)\text{ and }f(x)=y. $$ On $(-\infty,x_1)$, let $f(x)=y_1$, and on $(x,+\infty)$, let $f(x)=y$.

Answer 2

Here's a generalization: Let $X$ be a metric space, $x\in X,$ and $x_n$ is a sequence of distinct points in $X\setminus \{x\}$ converging to $x.$ Let $y,y_n$ be as before. Then there exists a continuous $f:X\to \mathbb R$ such that $f(x_n)=y_n, n=1,2,\dots$ and $f(x)=y.$

Proof (sketch): Let $E=\{x_n\}\cup \{x\}.$ Then $E$ is closed in $X.$ Define $g:E \to \mathbb R$ by setting $g(x_n)=y_n, n=1,2,\dots$ and $g(x)=y.$ Then $g$ is continuous on $E.$ By Tietze's extension theorem, there exists a continuous $f:X\to \mathbb R $ such that $f=g$ on $E.$ That does it.