For a function $f : D \rightarrow \mathbb{R}$ and a point $x_{0}$ in
$D$, define $A := \{x \text{ in } D \text{ } \mid x \geq x_{0}\}$ and
define $B := \{x \text{ in } D \text{ } \mid x \leq x_{0}\}.$ Prove
that $f : D \rightarrow \mathbb{R}$ is continuous at the point $x_{0}$
if and only if $f : A \rightarrow \mathbb{R}$ and $f : B \rightarrow
> \mathbb{R}$ are continuous at $x_{0}$.
First of all, here is the definition of continuity that I am referring to. Note that I am not using the typical $\epsilon-\delta$ definition of continuity but rather the sequential definition.
Definition: A function $f : D \rightarrow \mathbb{R}$ is said to be continuous at the point $x_{0}$ in $D$ provided that whenever
$\{x_{n}\}$ is a sequence in $D$ that converges to $x_{0}$, the image
sequence $\{f(x_{n})\}$ converges to $f(x_{0})$.
My attempt:
First, suppose $f : D \rightarrow \mathbb{R}$ is continuous at the point $x_{0}$. Then, let the sequence $\{x_{n}\}$ in $D$ converge to $x_{0}$. Per our assumption, it follows that $\{f(x_{n})\}$ converges to $f(x_{0})$. By the definition of our sets $A$ and $B$, we know that $x_{0} \in A$ and $x_{0} \in B$. Let $\{a_{n}\}$ be a sequence in $A$ that converges to $x_{0}$ and let $\{b_{n}\}$ be a sequence in $B$ that converges to $x_{0}$. Now, all I need to do is show $\{f(a_{n})\}$ converges to $f(x_{0})$ and $\{f(b_{n})\}$ converges to $f(x_{0})$, but I've tried for very long and cannot come up with a way to do this.
For the converse, we suppose $f : A \rightarrow \mathbb{R}$ and $f : B \rightarrow \mathbb{R}$ are continuous at the point $x_{0}$. Again, I would really appreciate some help with this part.
This is not a homework question. I am just working through a textbook, and I don't have any solutions or hints. I am trying to learn analysis on my own, and any sort of guidance would truly be appreciated.
Best Answer
Let $D\subseteq\mathbb{R}$, with $D\ne\varnothing$, and let $f : D \rightarrow \mathbb{R}$.
Fix $x_0\in D$.
Let $A=\{x\in D\mid x\ge x_0\}$, and let $B=\{x\in D\mid x\le x_0\}$.
First suppose $f$ is continuous at $x_0$.
Our goal is to show that the restriction of $f$ to $A$ and the restriction of $f$ to $B$ are both continuous at $x_0$.
Let $(a_n)$ be an arbitrary sequence of elements of $A$ such that $(a_n)$ approaches $x_0$.
Since $f$ is continuous at $x_0$, and the sequence $(a_n)$ approaches $x_0$, it follows that the sequence $(f(a_n))$ approaches $f(x_0)$.
Hence the restriction of $f$ to $A$ is continuous at $x_0$.
By analogous reasoning, the restriction of $f$ to $B$ is continuous at $x_0$.
Thus, the restriction of $f$ to $A$ and the restriction of $f$ to $B$ are both continuous at $x_0$, as was to be shown.
Conversely, suppose the restriction of $f$ to $A$ and the restriction of $f$ to $B$ are both continuous at $x_0$.
Our goal is to show that $f$ is continuous at $x_0$.
Let $(d_n)$ be an arbitrary sequence of elements of $D$ such that $(d_n)$ approaches $x_0$.
Let $J=\{j\mid d_{n_j}\in A\}$, and let $K=\{k\mid d_{n_k}\in B\}$.
For $j\in J$, let $a_j=d_{n_j}$, and for $k\in K$, let $b_k=d_{n_k}$.
Clearly, at least one of the sets $J,K$ is infinite.
First suppose $J,K$ are both infinite.
Since $(a_j)$ and $(b_k)$ are infinite subsequences of $(d_n)$, it follows that both subsequences converge to $x_0$.
Then, since the restriction of $f$ to $A$ and the restriction of $f$ to $B$ are both continuous at $x_0$, it follows that the subsequences $(f(a_j))$ and $(f(b_k))$ of the sequence $(f(d_n))$ converge to $f(x_0)$.
Fix $\epsilon > 0$.
Let $N$ be a positive integer sufficiently large such that
If $n\in \mathbb{N}$, then $n=n_j$ for some $j\in J$, or $n=n_k$ for some $k\in K$ (inclusive or).
Hence $n > N\implies |f(d_n)-f(x_0)| < \epsilon$.
It follows that the sequence $(f(d_n))$ converges to $f(x_0)$.
Next suppose one of $J,K$ is finite, $J$ say.
Then some tail of the sequence $(d_n)$ consists of terms which are all in $B$, hence, since the sequence $(d_n)$ converges to $x_0$, and the restriction of $f$ to $B$ is continuous at $x_0$, it follows that the sequence $(f(d_n))$ converges to $f(x_0)$.
Thus, in both cases, the sequence $(f(d_n))$ converges to $f(x_0)$.
It follows that $f$ is continuous at $x_0$, as was to be shown.