Continuous function show a bound on the image gives a closed preimage

Question

I'm trying to show that given f: $\mathbb{R}^2 \rightarrow \mathbb{R}$ continuous, $\{ (x,y) \in \mathbb{R}^2 | 0 \leq f(x,y) \leq 1 \}$ is closed. It seems to me that the Maximum-Minimum Theorem is at play here but I can't quite figure it out.

Answer 1

Say, $(x_{n},y_{n})\in\{0\leq f\leq 1\}$ such that $(x_{n},y_{n})\rightarrow(x,y)$, then $f$ being continuous implies that $f(x_{n},y_{n})\rightarrow f(x,y)$. But $f(x_{n},y_{n})\leq 1$, applying basic limit rule we have $f(x,y)\leq 1$, similar to that $f(x,y)\geq 0$, so $(x,y)\in\{0\leq f\leq 1\}$, so the set is closed.

Answer 2

Recall that a function $f\colon X\to Y$ is continuous if and only if the preimage $f^{-1}(A)$ of any closed set $A\subset Y$ is closed in $X$. Now apply this to the closed set $[0,1]\subset \mathbb R$.

Answer 3

An $\epsilon-\delta$ approach:

Let us denote $C:=\{(x,y)\in\mathbb{R}^2\vert 0\leq f(x,y)\leq1\}$ we will show that the complement of $C$, which I will mark as $A$, is open.

consider a point $a_0\in A$. since $A$ is the complement of $C$, we have $f(a_0)\notin[0,1]$. i.e either $f(a_0)>1$ or $f(a_0)<0$. In the first case we can take $\epsilon=f(a_0)-1$ and in the second $\epsilon=0-f(a_0)$.
We now use the continuity of $f$ to find a $\delta$ such that for every $a\in B_\delta(a_0)$ we have $|f(a)-f(a_0)|<\epsilon$. By our definition of $\epsilon$ we can conclude that $f(a)\notin[0,1]$, and so $a\in A$.
Since this holds for any arbitrary point $a$ within the ball we can conclude that $a_0\in B_\delta(a_0)\subset A$.

Finally, we have shown this for an arbitrary point $a_0$ within $A$, hence $A$ is open, and by definition, $C$ is closed.