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let $f:\mathbb R^n\longrightarrow \mathbb R$ a continuous map and $\alpha\in \mathbb R$ . let $$A=\{(x_1,\cdots,x_n)\in \mathbb R^n\;|\; f(x_1,\cdots,x_n) < \alpha\}$$ How to show that $A$ is open in $\mathbb R^n$ and to what extent this result can be generalized?

I know that the complement $A^c=\bigsqcup_{k\geq \alpha} B_k$ where $$B_k=\{(x_1,\cdots,x_n)\in \mathbb R^n\;|\; f(x_1,\cdots,x_n) =k\}$$ and that $B_k$ is closed because $B_k=f^{-1}(\{k\})$ but we can't deduce that $A^c$ is closed since this is not a FINITE union.

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just a typo: you probably mean "..deduce that $A^c$ is closed since.." –  student Feb 15 '12 at 15:09
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2 Answers

Hints:

1) The preimage of an open set under a continuous map is open.

2) Look at the set $[\alpha,\infty)$ instead of $\{k\}$.

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ok so $A=f^{-1}\{(-\infty,\alpha)\}$ and since $(-\infty,\alpha)$ is open then so is $A$. Can we generalize this for some more general space other than $\mathbb R^n$ –  palio Feb 15 '12 at 14:47
    
Yes, the fact that the domain of $f$ is $\mathbb R^n$ is irrelevant for this argument. –  Rasmus Feb 15 '12 at 16:06
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Here $A= f^{-1}( (-\infty,a) ) .$ It can be shown that for maps $ f:\mathbb{R}^n \to \mathbb{R}^m $ the $\epsilon - \delta$ definition of continuity is equivalent to the statement that for every set $O$ open in $ \mathbb{R}^m $, the preimage $f^{-1} ( O) $ is open in $\mathbb{R}^n.$ Since $ (-\infty, a) $ is open in $\mathbb{R}$ and $f$ is continuous, $A$ is also an open set.

This formulation of continuity in terms of open sets becomes the definition in more general spaces where open sets are still defined but concepts of distance between points is not. Consider the generalized definition:

Let $f: X \to Y $ be a map between topological spaces $X$ and $Y.$ Then $f$ is called continuous if for every set $O$ open in $Y$ we have $f^{-1} (O)$ open in $X.$

Thus if another similar problem gives again $f$ is continuous and $A$ is the preimage of an open set, we can automatically conclude $A$ is open.

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