Check if this set is compact: $\{ (x,y,z) \in \mathbb R^3 \mid |x| \le 1 \land y = 0 \land 0 \le z \le |x| \}$

Question

I need to check if S = $$\{ (x,y,z) \in \mathbb R^3 \mid |x| \le 1 \land y = 0 \land 0 \le z \le |x| \}$$ is compact.
I tried to visualize this set and - doing so - i realized that it's enough to use two dimensions for doing - this set looks like two symmetrical right triangles with legs $1$ and $1$, therefore one can infer that the set is bounded.
As for being closed:
$$S^c = \{(x,y,z) \in \mathbb R^3 \mid |x| > 1 \lor y \ne 0 \lor z < 0 \lor z >|x| \}$$
This set seems to be open (I do not know how to prove it formally) and so $S$ is closed, therefore, $S$ is compact.

Is my reasoning correct?

Answer 1

Your reasoning is quite correct. Lets fill gaps

therefore one can infer that the set is bounded.

Sure, but that's not a formal proof. Note that you have three conditions on coordinates which imply that

$$0\leq x\leq 1$$ $$y=0$$ $$0\leq z\leq |x|\leq 1$$

In particular $S\subseteq [0,1]\times\{0\}\times [0,1]\subseteq\mathbb{R}^3$ and therefore it is bounded.

This set seems to be open (I do not know how to prove it formally)

Showing that $S$ is closed or $S^c$ open is pretty much the same job, so lets stick to showing directly that $S$ is closed. Consider four functions:

$$f,g,h,k:\mathbb{R}^3\to\mathbb{R}$$ $$f(x,y,z)=|x|$$ $$g(x,y,z)=y$$ $$h(x,y,z)=z$$ $$k(x,y,z)=|x|-z$$

And note that

$$S=f^{-1}\big((-\infty, 1]\big)\ \cap\ g^{-1}\big(\{0\}\big)\ \cap\ h^{-1}\big([0,\infty)\big)\ \cap\ k^{-1}\big([0,\infty)\big)$$

and since each function is continous then... ?