Closed and connected graph of a function implies continuous function

Question

Let $f:\mathbb R \rightarrow \mathbb R$ be a function such that its graph is closed and connected. I am to prove that it is a continuous function. If I take a sequence $(x_n)$ such that $(x_n)$ converges to $x$ as $n$ goes to $\infty$ , I have to show that $f(x_n)$ converges to $f(x)$. Since the graph is closed the only thing that I have to show is that $f(x_n)$ converges to some point. Now, what do I do? Do I assume the contrary and use a divergent subsequence $f(x_{n_k})$ and intermediate value property to arrive at some contradiction?

Answer 1

Suppose $x_{n}\rightarrow x$ and $\left\vert f(x_{n})\right\vert \rightarrow \infty .$ We claim that there is a $\delta >0$ such that $% \left\vert y-x\right\vert \leq \delta \Rightarrow \left\vert f(x)-f(y)\right\vert <1$ or $\left\vert f(x)-f(y)\right\vert >2.$ If the claim is false then we can find a sequence $\{u_{n}\}$ converging to $x$ such that $1\leq \left\vert f(x)-f(u_{n})\right\vert \leq 2$ $\forall n.$ There is a subsequence $\{f(u_{n_{k}})\}$ of $\{f(u_{n})\}$ converging to some point $w.$ Since the graph is closed we get $w=f(x).$ But $1\leq \left\vert f(x)-w\right\vert .$ This proves the claim. Let $G$ be the graph of $f.$ Then $G\cap \{[a,b]\times %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion \}=(G\cap \{[a,b]\times %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion \}\cap \{(t,s):\left\vert f(x)-s\right\vert <1\})\cup (G\cap \{[a,b]\times %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion \}\cap \{(t,s):\left\vert f(x)-s\right\vert >1\})$ where $[a,b]$ is the interval $[x-\delta ,x+\delta ].$ If we prove that $G\cap \{[a,b]\times %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion \}$ is connected we get a contradiction because the two sets on the right contain $(x,f(x))$ and $(x_{n},f(x_{n}))$ for $n$ sufficiently large. This would prove that $x_{n}\rightarrow x$ implies that $\{f(x_{n})\}$ is bounded and the fact that $g$ is closed shows the only limit point of this bounded sequence is $f(x).$ It follows that $f$ is continuous. To complete the proof we prove that $G\cap \{[a,b]\times %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion \}$ is connected. If $g:G\cap \{[a,b]\times %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion \}\rightarrow \{0,1\}$ is continuous we can extend it to a continuous function $g:G\rightarrow \{0,1\}$ by making it constant on $g:G\cap \{(-\infty ,a]\times %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion \}$ and on $g:G\cap \{[b,\infty )\times %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion \}.$ The extended function must be a constant and so must be the original function.