证明或反驳ℝ²中集合Ω的开性、闭性与有界性

Let's walk through the key topological properties of this subset of ℝ² (using the standard Euclidean metric):

Is Ω an open set? #

Nope. Here's why: pick any point on the upper or lower edge of the set, say ((x₁, 1)). For a set to be open, every point must have some small open ball around it that's entirely contained within the set. But no matter how small you make the radius of that ball around ((x₁,1)), it will include points where (x₂ > 1)—and those points aren't in Ω. Since we can't find such a ball for these edge points, Ω can't be open.

Is Ω a closed set? #

Yes. A set is closed if it contains all of its boundary points. The boundary of Ω is exactly the lines (x₂ = 1) and (x₂ = -1) (any neighborhood around points on these lines will have points both inside and outside Ω). Since Ω includes all points where (|x₂| ≤ 1), those boundary points are definitely part of the set. For a more formal proof, you could note that the complement of Ω (points where (|x₂| > 1)) is open—any point in the complement has a small ball around it that stays in the complement—so by definition, Ω is closed.

Is Ω a bounded set? #

No way. A bounded set has to fit entirely inside some finite-radius ball centered at the origin. But look at points like ((n, 0)) where (n) is any positive integer—these are all in Ω, and their Euclidean norm is (n), which can get as large as we want. There's no finite number that can bound the norm of every point in Ω, so it's unbounded.

内容的提问来源于stack exchange，提问作者hologram