# Proving the existence of a continuous function that satisfy a certain property from a finite version of this property

Let $M \subseteq [0,1] \times \mathbb{R}^n$ be a compact semialgebraic set. In particular, $M$ can be described by a finite set of polynomial equalities and inequalities. Let $\delta_0 > 0$ be a real number.

Suppose that the following condition holds: for every finite collections $t_1,t_2,\cdots,t_K \in [0,1]$ there is a continuous function $x : [0,1] \to \mathbb{R}^n$ that satisfies:

1) The graph of $x$ is a subset of $M$.

2) $\|x(t_k)\| \geq \delta_0$ for every $k\in \{1,2,\ldots,K\}$.

I would like to deduce that there is a continuous function $x : [0,1] \to \mathbb{R}^n$ that satisfies:

1) The graph of $x$ is a subset of $M$.

2) $\|x(t)\| \geq \delta_0$ for every $t \in [0,1]$.

Is this true?

• A constant function $x$ is not what you want, right?
– Dirk
Commented Nov 8, 2017 at 10:05
• Dear Dirk, I would not mind having a constant function, as long as its graph is a subset of M (and the constant is above $\delta_0$). Commented Nov 8, 2017 at 12:00
• Oh, I missed that $M\subseteq [0,1]\times \mathbb{R}^n$ (somehow I assumed that $M\subseteq \mathbb{R}^n$…)
– Dirk
Commented Nov 8, 2017 at 14:38
• That is obviously false as written now: take $n=1$ and $M=\{(y-x+0.5)^2\le 0.01\}$. Then at every particular point you have an interval of length $0.2$ of admissible values, so you can choose a value of size at least $0.1$ and extend linearly between the points but your $f$ has to change sign on $[0,1]$, so the intermediate value theorem is a killer. I surmise you meant something else. Commented Nov 9, 2017 at 4:25
• Thanks. In my original question I had $\mathbb{R}^n_+$ instead of $\mathbb{R}^n$ throughout, and I thought that changing this to $\mathbb{R}^n$ would not matter. In fact, it does not, since my son came up with a counterexample to the case where we require $M$ to be a subset of the nonnegative orthant: take $M = (t,x,(t-\frac{1}{2})^2)$, where $t,x \in [0,1]$ and $\frac{x}{3} \leq t \leq \tfrac{2+x}{3}$. Commented Nov 9, 2017 at 21:04