# Continous map assuming positive value in the closure of a convex set

Let $\phi:\mathbb{R}^d\rightarrow\mathbb{R}$ be a continuous map w.r.t. Euclidean norm. Let $C\subseteq\mathbb{R}^d$ be a convex subset. Assume that there exists $y\in\overline{C}$ such that $\phi(y)>0$. Does there exist $y'\in C$ such that $\phi(y')>0$? Of course, if $y\in C$ then we are done. But what if $y\in\overline{C}\setminus C$? Maybe convexity is not needed here, but in my case, convexity is given.

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Since $\phi(y)>0$, there is an interval around it, say $(\delta,\varepsilon)$, with $0<\delta<\varepsilon$. Now, since $\phi$ is continuous, $\phi^{-1}(\delta,\varepsilon)$ will be open, and in particular, we can take the component containing $y$ and see that it intersects $C$. This implies there is an element $y'\in C$ such that $\phi(y')\in(\delta,\varepsilon)$. Equivalently, $\phi(y')>0$.
why can't it be that $\phi^{-1}(\delta,\varepsilon)\cap C=\emptyset$? – Andy Teich Mar 11 '13 at 13:23
One of the equivalent definitions of $\overline{C}$ is this: if $y\in\overline{C}$, then any open set containing $y$ contains a point $y'$ in $C$ distinct from $y$. Note that this is where the convexity is required (perhaps less, I think you can do it with connected). If $C$ was the unit ball unioned with $(2,2,2)$, then the fact that $\phi(2,2,2)>0$ has no bearing on the unit ball. – Clayton Mar 11 '13 at 13:24
@AndyTeich: Which part has you confused? The characterization I gave is indeed for boundaries, and so we somehow have to exclude bad points (like the $(2,2,2)$ above). In order to do this, we impose some niceness condition on the set, say convexity or connectedness. – Clayton Mar 11 '13 at 13:46