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I am trying to show that $f(a,b,c):R^3\to R\;$ defined by $f(a,b,c) = a^2+b^2+c^2+1$ is contunious by showing that for every open set, $U$, in $R$ has an open preimage - $f^{-1}(U)$ is open in $R^3$.

All I really have so far is that as $U$ is open, we know that:

$\forall a \in U, \; \exists \varepsilon>0$ such that $N(a,\varepsilon)\subset U$

From here I'm super stuck. I think I want to get to:

$\forall (x,y,z) \in f^{-1}(U) \; \; \exists \varepsilon^{\prime}$ such that $N((x,y,z),\varepsilon^{\prime})\subset f^{-1}(U)$

And I think to do so, I will need to use the function itself, but I am not sure how to do so.

Any help would be greatly appreciated. Thank you!

Ruy
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TNoms
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1 Answers1

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Take $U = (q-\epsilon,q + \epsilon)$. Then, $f^{-1}(U) = \{ (a,b,c) : q - \epsilon < a^2 + b^2 + c^2 + 1 < q + \epsilon \}$. That condition becomes $$ q - 1 - \epsilon < a^2 + b^2 + c^2 < q -1 + \epsilon. $$ This set describes either an open ball ($q - 1 - \epsilon < 0$) or an open ball minus a closed ball ($q - 1 + \epsilon \geq 0$). Either way, we obtain that $f^{-1}(U)$ is open.

Since it's possible to write any open subset of $\mathbb{R}$ as a union over sets of the above form, you get that $f^{-1}(U)$ is open for any $U \subseteq \mathbb{R}$ open.

Wizact
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