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Let $f$ be a continuous real-valued function defined on an open subset $U$ of $\mathbb{R}^n$.

Show that $\{(x,y):x\in{U},y>f(x)\}$ is an open subset of $\mathbb{R}^{n+1}$


Let $\forall{x}\in{X}, X\subset{U}$

Using the theorem, for a function $f$ mapping $S\subset{\mathbb{R}^n}$ into $\mathbb{R}^m$, it is equivalent to $f$ is continuous in $S$

so we can say $f(x)$ is continuous on $U$. Also, by following $U$, $f(x)$ is also open which is one of what I want to prove.

But how does it so sure about it maps to $\mathbb{R}^{n+1}$ but not $\mathbb{R}^{n+2}$, $\mathbb{R}^{n+3}$, ...

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Hint: the function $$ g:(x,y)\longmapsto y-f(x) $$ is defined and continuous on $U\times \mathbb{R}$, which is open in $\mathbb{R}^{n+1}$. Now try to express your set with this function.

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Why not $\mathbb{R}^n$??? –  Paul Mar 11 '13 at 0:03
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@Paul What do you mean? The function $f$ is defined on $U$, so the natural domain of this function is $U\times \mathbb{R}$, that is the points $(x,y)$ for which the expression $y-f(x)$ makes sense. –  1015 Mar 11 '13 at 0:04
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