Show {$(x,y)\subset R^2|y

Question

$f$ is continuous on $[a,b]$,I want to show that {$(x,y)\subset R^2|y

Answer 1

Consider the map $d:(a,b)\times\mathbb{R}\to\mathbb{R},(x,y)\mapsto f(x)-y$.

$d$ is continuous, thus $d^{-1}((0,+\infty))$ is an open subset of $(a,b)\times\mathbb{R}$, hence an open subset of $\mathbb{R}^2$

Answer 2

You can also do this by showing that every point has a ball around it.

Take a point $(u, v)$ in $\{(x,y) \subset \mathbb{R}^2 \mid y

By the intermediate value theorem, there is some point in that ball which has $v_n = f(u_n)$. (Just take the line between $(u_n, v_n)$ and $(u,v)$, and project onto that line; then use the IVT.)

This gives a sequence in $\{(r,s): s = f(r)\}$ which tends to $(u,v)$, and that set is closed (as you noted, since it's the graph of a continuous function), so $(u,v)$ must be in the graph, a contradiction.