请求构造$R^3$中既非开、非闭、非凸也非有界的点集示例

Let's define the set ( S \subseteq \mathbb{R}^3 ) as:

S = { (x, y, z) ∈ ℝ³ | x² + y² < 1, z > 0 } ∪ { (n, 0, 0) | n ∈ ℕ }

Breaking this down: it's all points above the xy-plane that lie inside the open unit circle centered at the origin, plus all points along the x-axis at positive integer coordinates (1,0,0), (2,0,0), (3,0,0), etc.

Now let's verify each property one by one:

• Not open: Take the point (1, 0, 0) which is in S. Any open ball around this point will contain points like (1 + ε, 0, 0) where ε is a tiny positive number—these points aren't in S (they don't satisfy the circle condition, and 1+ε isn't a natural number). Since (1,0,0) is an element of S but not an interior point, S can't be an open set.

• Not closed: Consider the point (0, 0, 0). We can find a sequence of points in S that converges to (0,0,0) (for example, (0, 0, 1/k) for k = 1,2,3,...). But (0,0,0) isn't in S (since z=0 doesn't satisfy z>0, and it's not a natural-number x-coordinate point). Since S has a limit point not contained in itself, it's not closed.

• Not convex: Pick two points in S: (0, 0, 1) (from the open circle/upper half-space part) and (2, 0, 0) (from the integer x-axis points). Their midpoint is (1, 0, 0.5). This point doesn't satisfy x² + y² < 1 (since x=1 makes x²+y²=1), and it's not one of the integer x-axis points. So the midpoint isn't in S, which means S fails the convexity condition.

• Not bounded: The points (n, 0, 0) for n ∈ ℕ can be made arbitrarily far from the origin as n increases. There's no finite-radius ball that can contain all elements of S, so it's unbounded.

内容的提问来源于stack exchange，提问作者Sanskruti More