$f:\mathbb R \to \mathbb R^n$ be such that $G(f):=\{(x,f(x)):x \in \mathbb R\}$ is closed and connected in $\mathbb R^{n+1}$ , is $f$ continuous ?

Question

Let $f:\mathbb R \to \mathbb R^n$ be a function whose graph $G(f):=\{(x,f(x)):x \in \mathbb R\}$ is closed and connected

in $\mathbb R^{n+1}$ , then is $f$ continuous ?

Answer 1

Revamped. The claim is indeed true.

Suppose that $G(f)$ is closed and connected in $\mathbb{R}\times\mathbb{R}^n$. We augment $\mathbb{R}^n$ with the point at infinity, namely $\infty$. In other words, we consider the one-point compactification $\mathbb{R}^n \cup \{\infty\}$ of $\mathbb{R}^n$. (Note: The use of the notion of "point at infinity" here is merely for simplicity. Every statement below involving $\infty$ can be unpacked into a more elementary version.)

We begin by establishing:

Claim. Let $a \in \mathbb{R}$. Then, as $x \to a$,

1. the only possible limit points of $f(x)$ are $f(a)$ and $\infty$, and in fact,
2. $f(a)$ is always a limit point of $f(x)$.

Proof of Claim. (1) Assume otherwise, and let $b \in \mathbb{R}^n \setminus\{a\}$ be a limit point of $f(x)$ as $x \to a$. Then there exists a sequence $(x_k) \subseteq \mathbb{R}\setminus\{a\}$ such that $x_k \to a$ and $f(x_k) \to b$. Then $(x_k, f(x_k)) \to (a, b)$, hence by the closedness of $G(f)$, we have $b = f(a)$, a contradiction.

(2) Assume $f(a)$ is not a limit point of $f(x)$ as $x \to a$. Then by the previous step, $\infty$ is the only limit point, and in particular, we have $\lim_{x\to a} \|f(x)\| = +\infty$. Hence, there exist $\delta > 0$ such that the closed neighborhood, $$ K = [a-\delta, a+\delta] \times \{ y \in \mathbb{R}^n : \|y - f(a)\| \leq 1\}, $$ satisfies $G(f) \cap K = \{(a, f(a))\}$. Then the open sets $U = \mathring{K}$ and $V = K^c$ altogether separate the graph $G(f)$, contradicting the connectedness assumption.

So, it suffices to show that any of the following equivalent statements hold:

1. $\infty$ cannot be a limit point of $f(x)$ about any point in $\mathbb{R}$.
2. $f(x)$ is locally bounded.

Assume otherwise, so that there exists $a \in \mathbb{R}$ such that $\infty$ is a limit point of $f(x)$ as $x \to a$. We show that this implies a contradiction that $G(f)$ is disconnected.

Indeed, by the above lemma, the limit points of $f(x)$ as $x \to a$ are precisely $f(a)$ and $\infty$. So by choosing disjoint neighborhoods of $f(a)$ and $\infty$, respectively, and then invoking the epsilon-delta definition, we can find $p, q$ with $p < a < q$ such that the graph of $f$ restricted on $[p, q]$,

$$ G(f|_{[p, q]}) = \{(x, f(x)) : x \in [p, q]\}, $$

is disconnected. Consequently, there exists a surjective continuous function

$$ \phi : G(f|_{[p, q]}) \to \{0, 1\} $$

that "separates" $G(f|_{[a, b]})$. Now we extend $\phi$ to all of $G(f)$ by:

$$ \phi((x, f(x))) = \begin{cases} \phi((p, f(p))), & \text{if $x \leq p$}, \\ \phi((x, f(x))), & \text{if $x \in [p, q]$}, \\ \phi((q, f(q))), & \text{if $x \geq q$}. \end{cases} $$

In other words, we extend $\phi$ so that it is constant either on $(-\infty, p] \times \mathbb{R}^n$ or on $[q, \infty) \times \mathbb{R}^n$. Then $\phi$ is still a surjective continuous function, contradicting the connectedness of $G(f)$ as required. $\square$

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Disclaimer. The idea of using separating function in the last part of the proof is borrowed from @Kavi Rama Murthy's answer.