Show closedness , path connectedness and compactness

Question

Let f : R2 →R be a continuous function. Let S = {(x,y,z): z = f(x,y)}. Show that S is closed, path connected but not compact.

Unfortunately inspite of knowing the definitions, I donot know how to prove

Answer 1

Hints:

• Look at $F\colon \Bbb R^3 \to \Bbb R$ given by $F(x,y,z) = z-f(x,y)$. Is $F$ continuous? What is $F^{-1}[\{0\}]$?

• Given two points $(x_1,y_1,f(x_1,y_1)),(x_2,y_2,f(x_2,y_2)) \in S$, can you at least join $(x_1,y_1)$ to $(x_2,y_2)$ in $\Bbb R^2$? Can you "lift" that path to $S$? How?

• Is $\{(n,0,f(n,0)) \mid n \geq 1\} \subseteq S$ bounded?

Answer 2

For closedness, let $(x_n,y_n,z_n) \rightarrow (x,y,z)$ with $(x_n,y_n,z_n) \in S$ for all $n$ and note that $f(x_n,y_n) \rightarrow f(x,y)$. For path-connectedness, fix $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in $S$, let $g: [0,1] \rightarrow \mathbb{R}^2, t \mapsto t(x_1,y_1)^T+(1-t)(x_2,y_2)^T$ and consider $h: [0,1] \rightarrow S, t \mapsto (g(t), f(g(t)))$. To disprove compactness, consider the sequence $(n,n,f(n,n))$.