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(1)for $(x,y,z) \in \, \mathbb{R}^3, \quad x ^2 + y ^2 < 1, \quad z =0$

judge if this is open, closed, and bounded or not. and give proofs.

I'm not sure whether it's open or not. I think it's open, not closed and bounded. Because whatever neighbouhood you choose, it would certainly contain other points since z is always 0.then if you choose a open ball centered at any point, it would be 3 dimensional and go beyond the 2-d shape.

(2) judge whether a finite set is bounded, closed, open or not.Is finite set just some finite points?

I'm not sure about the answer either.

What do you guys think?

2 Answers 2

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Your argument shows that in case $(1)$, the set is not open. It isn't closed, either (since the boundary is $x^2+y^2=1,z=0$), but it is bounded.

A finite point set is closed (since it has no limit points) and bounded (since the points' distances from the origin have a maximum), and isn't open unless it's empty.

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$S$ is bounded, not open in $\mathbb{R}^3$. It is only relative open.

See here

http://en.wikipedia.org/wiki/Relative_interior

for definition of relative interior.