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I've been studying the implicit and inverse functions theorems and I've started with one special case of the implicit function theorem. The book I'm reading states the theorem as follows:

Let $f : U\to\mathbb{R}$ be a function of class $C^k$ with $k\geq 1$ defined on some open set $U\subset\mathbb{R}^n\times\mathbb{R}$. If $p=(x_0,y_0)\in U$ is such that $f(p)=c$ and $D_{n+1}f(p)\neq0$, then there exists a ball $B(x_0;\delta)\subset \mathbb{R}^n$ and one interval $J=(y_0-\epsilon,y_0+\epsilon)$ such that $f^{-1}(c)\cap (B\times J)$ is the graph of a function $\xi : B(x_0;\delta)\to J$ of class $C^k$ and for all $x\in B(x_0;\delta)$ we have $D_i\xi(x)=-D_if(x,\xi(x))/D_{n+1}f(x,\xi(x))$.

Now, the proof starts as follows:

Consider that $D_{n+1}f(x_0,y_0)>0$, then since $D_{n+1}f$ is continuous, there exists $\delta>0$ and $\epsilon>0$ such that if $B$ is the open ball with center $x_0$ and radius $\delta$ and if $J=(y_0-\epsilon,y_0+\epsilon)$ then $B\times \bar{J}\subset U$ and $D_{n+1}f(x,y)>0$ for all $(x,y)\in B\times \bar{J}$.

My doubt is exactly there. First: why can we assure that $B\times\bar{J}\subset U$ based on continuity? I know that if some continuous function is positive in some point, then there's a neighbourhood of the point where it is still continuous, but why can we take the neighbourhood like that? That closure confused me a little.

The only thing I could think of was the following: if we endow $\mathbb{R}^n\times\mathbb{R}$ with the product topology and endow $\mathbb{R}^n,\mathbb{R}$ with the metric topology, then a set of $\mathbb{R}^n\times\mathbb{R}$ is open if and only if for every point of the set there's one product $U_1\times U_2$ of open sets $U_1\subset\mathbb{R}^n,U_2\subset\mathbb{R}$. Now, since $D_{n+1}f$ is continuous and $D_{n+1}f(p)>0$, then there's a neighbourhood $N(p)\subset U$ such that for every $q\in N(p)$ we have $D_{n+1}f(q)>0$. Since it is a neighbourhood of $p$, there must be inside of it an open set containing $p$, now we can take a basic one, because the definition of the open sets. But why can we assure that the product of the ball with the closure of the interval is still in $N(p)$?

Thanks very much in advance!

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~~Edited out unneeded conversation about product topology ~~

If $B\times J \subset U$ but $B\times \overline J$ isn't in $U$, then just replace $J$ by $J' = (y_0 - \epsilon/2,y_0 + \epsilon/2)$.

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    $\begingroup$ But I've said: "a set of $\mathbb{R}^n\times\mathbb{R}$ is open if and only if for every point of the set there's one product $U_1\times U_2$ of open sets $U_1\subset\mathbb{R}^n, U_2\subset \mathbb{R}$", isn't that the way we define a topology generated by a basis? At every point of the open set there's a basic open set containing the point still inside the set? Now I've got the thing with the closure! Thanks very much @BaronVT! $\endgroup$ – user1620696 Oct 8 '13 at 19:23
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    $\begingroup$ You're welcome! And you're right about the basis, I just didn't read what you wrote carefully enough... $\endgroup$ – BaronVT Oct 8 '13 at 19:33

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