Given locally limited $f:[0,1]→\mathbb{R}$, show that $Graph(f)$ is closed in $\mathbb{R^2}$ ⟺ $f$ is continuous using sequences

Note that $f$ being locally limited means that given $x∈[0,1]$ there is an open interval $(a,b)$ in $\mathbb{R}$, that contains $x$, such that $f_{|(a,b)∩[0,1]}$ is limited.

Usual metric.


Let $G$ be the graph of $f$. Suppose $f$ is continuous. Let the sequence $(x_n,f(x_n))\in G$ converge to $(x,y)$. We need to show that $y=f(x)$ so that $(x,y)\in G$. But $$ y=\lim f(x_n)=f(\lim x_n)=f(x)$$ Now suppose $G$ is closed. Suppose $x_n \to x$. We need to show that $f(x_n)\to f(x)$. If this does not happen then there is a subsequence $x_{n_k}$ such that $|f(x_{n_k})-f(x)|\ge \epsilon $ for some $\epsilon >0$. But for large enough $n_k$ the $x_{n_k}$ is close enough to $x$ so $f(x_{n_k})$ is bounded. And as a bounded sequence it must have a convergent subsequence $f(x_{n_{k_j}})\to y$. So the sequence $(x_{n_{k_j}},f(x_{n_{k_j}}))$ in $G$ converges to $(x,y)$. Therfore we must have $y=f(x)$. So $f(x_{n_{k_j}})\to f(x)$. But this contradict the fact that $|f(x_{n_{k_j}})-f(x)|\ge \epsilon$. So $f(x_n)$ must converge to $f(x)$ and $f$ is continuous.

The answer is accepted.
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