Let A = [0, 1] ⊆ ℝ with the standard topology. Is the set [0, 1/2) open in A with the subspace topology?
ANo — 0 is a boundary point of [0, 1/2) in ℝ, so it cannot be open in any topology
BYes — because [0, 1/2) = (−1, 1/2) ∩ [0, 1], and (−1, 1/2) is open in ℝ
CNo — open sets in A must also be open in ℝ
DYes, but only if A is given the discrete topology
The subspace topology on A consists of all sets of the form U ∩ A where U is open in ℝ. The set (−1, 1/2) is open in ℝ, and (−1, 1/2) ∩ [0, 1] = [0, 1/2). So [0, 1/2) is open in A. The key insight: openness is *relative to the space you are working in*. Within A's own topology, the point 0 has an open neighborhood [0, ε) in A — it is surrounded by open sets in A. Option A is the common confusion: it evaluates openness from ℝ's perspective rather than A's. The subspace topology explicitly allows A to have open sets that contain 'boundary' points of A as seen from ℝ.
Question 2 Multiple Choice
The subspace topology on A ⊆ X is characterized by which universal property?
AIt is the largest topology on A making the inclusion map ι: A → X continuous
BIt is the smallest (coarsest) topology on A making the inclusion map ι: A → X continuous — and a map f: B → A is continuous if and only if the composition ι ∘ f: B → X is continuous
CIt ensures every subset of A is open, making any map into A continuous
DIt copies every open set of X onto A without modification
The subspace topology is the *initial topology* with respect to the inclusion map — the coarsest (fewest open sets) topology on A making ι continuous. It cannot be made coarser without breaking the continuity of ι. The universal property states that to check continuity of a map f: B → A, it is equivalent to check that ι ∘ f: B → X is continuous. This is why the subspace topology is 'right': it is the natural choice forced by requiring the inclusion to be continuous. Option A has the direction backwards — the largest topology on A would be discrete, which gives ι: A → X continuity trivially but imposes more open sets than necessary.
Question 3 True / False
A subset C ⊆ A is closed in A (with the subspace topology) if and only if C = F ∩ A for some set F that is closed in X.
TTrue
FFalse
Answer: True
This follows directly from the definition. A set is closed if its complement is open. C is closed in A iff A \ C is open in A iff A \ C = U ∩ A for some open U in X. Taking complements: C = A \ (U ∩ A) = A ∩ (X \ U). Let F = X \ U, which is closed in X. Then C = F ∩ A. The same structure that defines open sets in A — intersecting open sets of X with A — gives an analogous description of closed sets.
Question 4 True / False
If U is open in the subspace topology on A ⊆ X, then U is expected to also be open in X.
TTrue
FFalse
Answer: False
This is the key subtlety of subspace topology: open in A does not imply open in X. A = [0, 1] with the subspace topology from ℝ: the set [0, 1/2) is open in A (it is (−1, 1/2) ∩ [0, 1]) but is not open in ℝ — it contains 0, which has no open interval around it lying entirely in [0, 1/2). Openness is always relative to a specific topology on a specific space. When we say U is open in A, we mean U belongs to the subspace topology τ_A, not that U belongs to the original topology τ on X.
Question 5 Short Answer
Explain why, in the subspace topology on A = [0, 1] ⊆ ℝ, the endpoint 0 is an interior point of A even though it is a boundary point of [0, 1] when viewed from ℝ.
Think about your answer, then reveal below.
Model answer: Interior in the subspace topology means 0 has an open neighborhood *within A's topology* that contains only points of A. The set [0, ε) = (−ε, ε) ∩ [0, 1] is open in A for any ε > 0, and it is a neighborhood of 0 lying entirely within A. Since A only 'sees' its own topology — it does not look outside itself to ℝ — 0 is surrounded by open sets within A. The concept of interior is always relative: a point is interior if it has an open neighborhood in the topology of the space it belongs to. From ℝ's perspective, 0 is a boundary point of [0, 1] because any open interval around 0 in ℝ contains points outside [0, 1]. But once we restrict to A, those external points no longer exist in our space, and 0 gains interior status.