開集合、コンパクト性と連結性
導入済み| English | 意味 |
|---|---|
| connectedness/kəˈnektɪdnəs/ | 連結性 |
| relative topology/ˈrelətɪv təˈpɒlədʒi/ | 相対位相 |
A decision before an answer
- The interval [0,1) is open in one natural space and neither open nor closed in another. The surrounding space is part of the question.
- Your goal: Compute closure, interior and boundary in a stated space.
Read the relationship
- In a metric space, an open set contains a small ball around each of its points. The interior consists of such points; the closure includes all limit points; the boundary is closure minus interior. Open and closed are not mutually exclusive labels; the empty set and the whole space are both. In R, the set [0,1) has interior (0,1), closure [0,1] and boundary {0,1}.
- Apply compactness and connectedness to continuous maps.
What is the boundary of (0,1) as a subset of R?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The closure is [0,1] and the interior is (0,1); removing the latter leaves both endpoints.
Use the defining rule
- In a subspace X, an open set has the form X intersected with an ambient open set. Thus [0,1) is open relative to [0,2], using intersection with (-1,1), although it is not open in R. A set may also be relatively closed without being closed in the ambient space. Always state which space defines neighbourhoods and which metric is used.
- Distinguish relative topology from the ambient Euclidean topology.
Which subset of R is connected but not compact?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
An open interval is connected but not closed in R, so it is not compact. The other options are compact or disconnected.
Check the conditions
- Compactness means every open cover has a finite subcover. In Euclidean R^n, Heine–Borel makes this equivalent to closed and bounded. In a metric space, compactness is equivalent to sequential compactness; completeness and boundedness alone do not suffice in arbitrary metric spaces. A continuous image of a compact set is compact, giving attained maxima and minima for real continuous functions on a nonempty compact domain.
- Distinguish relative topology from the ambient Euclidean topology.
Let X=[0,1] with its usual relative topology. The set U=[0,0.5) equals X intersected with (-1,0.5), so U is open in X. Its closure in X is [0,0.5], and its boundary in X is {0.5}; zero is an interior point relative to X. For a continuous f on X with f(0)<0<f(1), connectedness ensures a zero, while compactness separately ensures attained extrema.
The number of boundary points of [0,0.5) relative to X=[0,1] is ____.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Only 0.5 is a relative boundary point. Zero has a relative neighbourhood contained in the set.
Apply the task format
- Connected sets cannot be separated into two disjoint nonempty relatively open parts; connected subsets of R are precisely intervals. A continuous image of a connected set is connected, which yields the intermediate value theorem. Path connectedness implies connectedness, but not conversely in every space. A compact set need not be connected, and a connected set need not be compact; a finite two-point set and an open interval supply the contrasting cases.
- Distinguish relative topology from the ambient Euclidean topology.
Heine–Borel's closed-and-bounded test is a Euclidean-space theorem. Connectedness and compactness answer different questions.
このケースに当てはまる答えはどれか？
指定された空間における閉包、内部および境界を計算する
Every continuous function on a bounded open interval attains a maximum.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
f(x)=x on (0,1) has supremum 1 but no maximum. Compactness of the domain would supply the missing hypothesis.
Keep the distinctions
- relative topology 相对拓扑 — Open subsets inherited by intersecting a subspace with ambient open sets.
- connectedness 连通性 — Absence of a separation into disjoint nonempty relatively open subsets.
- Compute closure, interior and boundary in a stated space.
- Apply compactness and connectedness to continuous maps.
- Distinguish relative topology from the ambient Euclidean topology.
Match each term with its precise meaning in this lesson.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
The order follows the stated process; check each stage before the next.