Let $f$ $\textbf{R}^n \rightarrow \textbf{R}^m$ be a continuous everywhere function and $c Prove {$x \in \textbf{R}^n | c<|f(x)| I tried to prove this by definition, i.e. a set is open iff every point in it has an open ball centered at the point.
So take any point $x$ in the set, it follows that $c<|f(x)|
Prove {$x \in \textbf{R}^n |c<|f(x)|
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real-analysis
general-topology
metric-spaces
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0You are not indicating exactly *how* you are using the definition of continuity to arrive at the conclusion. E.g., if this is supposed to be based on a $\varepsilon$-$\delta$ definition, what $\varepsilon$ do you need? A shorter proof if you have the background would be based on the following: $f$ continuous implies $|f|$ is continuous, and the inverse image of an open set under a continuous function is open. – 2017-02-28
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$$\{x\in\mathbb{R}^n\mid c<|f(x)|
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0You edited the question by $f:\mathbb{R}^n\to\mathbb{R}^m$. Thus you need to use the continuous function $|f|:\mathbb{R}^n\to \mathbb{R}$ which is continuous, being composition of two continuous function. – 2017-03-03