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Let $f:\mathbb{R}^n\rightarrow\mathbb{R}\;(n\in \mathbb{N})$ be a function. Suppose that $f$ is continuous at $\bar{x}$. Let $$ \text{epi}f:=\{(x, \lambda)\in \mathbb{R}^n\times\mathbb{R}: f(x)\leq \lambda\}. $$ I would like to ask whether $\text{epi}f$ is locally closed at $(\bar{x}, f(\bar{x}))$.

Thank you for all your helping.

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Consider $n = 1$ and the function \begin{align*} f \colon \mathbb R &\to \mathbb R\\ x &\mapsto \begin{cases} 1/q & x = p/q \in \mathbb Q, \operatorname{gcd}(p,q) = 1\\ 0 & x \not\in \mathbb Q\end{cases} \end{align*} Let $\bar x \in \mathbb R \setminus \mathbb Q$, then $f$ is continuous at $\bar x$. Let $U$ be a neighbourhood of $(\bar x, 0)$, then $U \supseteq (\bar x - \epsilon, \bar x + \epsilon) \times (-\epsilon, \epsilon)$ for some $\epsilon > 0$. Let $p/q \in (\bar x - \epsilon, \bar x + \epsilon)\cap \mathbb Q$, and $x_n \to p/q$ be a sequence of irrationals in $(\bar x -\epsilon, \bar x +\epsilon)$. Then $(x_n,0) \in \operatorname{epi} f\cap U$ for each $n$, but their limit $(p/q, 0)\in U \setminus \operatorname{epi} f$. Hence $\operatorname{epi} f$ is not closed in $U$. As $U$ was an arbitrary neighbourhood of $(\bar x, 0)$, it is not locally closed at $(\bar x, 0)$.

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Thank you for your solution. – blindman Oct 19 '12 at 9:24

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