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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

Didi
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    These sort of image sets are always closed since the set of points where two continuous functions are equal is closed. Lack of compactness and preservation of path connectedness follows from $\mathbb R^2$ not being compact but path connected and your map being continuous. – Brevan Ellefsen Nov 09 '19 at 20:18
  • A set is closed iff it contains all of its limit points. A set $S$ is path connected if for any two points $x,y\in S$, there exists a continuous function $f:[0,1]\to S$ with $f(0)=x$ and $f(1)=y$. A set is compact if every cover by open sets has a finite sub-cover. Do those definitions help? – Math1000 Nov 10 '19 at 00:28
  • Yes, thanks but can you please relate them with the question – Didi Nov 10 '19 at 04:47

2 Answers2

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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?

Ivo Terek
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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 $(n,n,f(n,n))$.

Nicolas
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