Consider the projection π : ℝ² → ℝ defined by π(x, y) = x, and the closed set C = {(x, y) : xy = 1} (the hyperbola). What is π(C) in ℝ, and what does this reveal about π?
Aπ(C) = ℝ, which is both open and closed — showing π is both open and closed
Bπ(C) = (−∞, 0) ∪ (0, ∞), which is open in ℝ — showing that π is continuous but not closed
Cπ(C) = ℝ \ {0}, which is closed in ℝ — showing π is a closed map
Dπ(C) = [−1, 1], since the hyperbola is bounded — showing π is neither open nor closed
The hyperbola C approaches but never reaches (0, y) for any y — as x → 0, y = 1/x → ∞. So the projection misses 0: π(C) = (−∞, 0) ∪ (0, ∞). This set is open in ℝ, not closed (0 is a limit point not in the set). This shows π fails to be a closed map — it sends a closed set to an open set. Yet π is continuous (a basic fact) and open (projections of open sets in ℝ² are open in ℝ). This example is the canonical demonstration that continuity and openness do not imply closedness.
Question 2 Multiple Choice
The map f : [0, 1) → S¹ that wraps the half-open interval onto the circle is a continuous bijection. Why is it not a homeomorphism?
AIt is not injective — two distinct points in [0, 1) map to the same point on the circle
BIt is not continuous — there is a discontinuity at the endpoint 0
CIts inverse is not continuous, which means f fails to be an open map — open sets in [0, 1) near 0 do not map to open sets in S¹
DEvery continuous bijection between compact spaces is automatically a homeomorphism, so this map must be a homeomorphism
The set [0, 0.5) is open in [0, 1) (in the subspace topology), but its image in S¹ includes the 'join point' of the circle without an open arc entirely surrounding it — so the image is not open in S¹. This means f is not an open map, and therefore its inverse is not continuous. The continuous+bijective combination is insufficient for a homeomorphism; you need also openness (or closedness). Option D is a common error: the theorem that continuous bijections from compact spaces to Hausdorff spaces are homeomorphisms requires compactness of the domain — [0, 1) is not compact.
Question 3 True / False
Continuity of f : X → Y is defined in terms of preimages of open sets, while openness of f is defined in terms of images of open sets.
TTrue
FFalse
Answer: True
This is the precise distinction between the two properties, and it explains why they are logically independent. Continuity: for every open V ⊆ Y, f⁻¹(V) is open in X. Openness: for every open U ⊆ X, f(U) is open in Y. The direction is reversed. A map can be continuous but not open (e.g., sin : ℝ → [−1, 1]), open but not continuous (a bijection between a discrete and an indiscrete space), both, or neither.
Question 4 True / False
A map that is both continuous and open is expected to be a homeomorphism.
TTrue
FFalse
Answer: False
A homeomorphism requires continuous + open + bijective. Continuous and open alone are not enough. The projection π : ℝ² → ℝ is both continuous and open, but it is not bijective (many points map to the same value), so it is not a homeomorphism. You need bijectivity as well. The equivalent characterization of a homeomorphism is: bijective, continuous, and either open or closed (since for a bijection, openness and closedness are equivalent).
Question 5 Short Answer
Give an example of a continuous map that is not an open map, and explain what structural feature prevents open sets from having open images.
Think about your answer, then reveal below.
Model answer: The function f : ℝ → ℝ defined by f(x) = x² is continuous but not open. The image of the open set (−1, 1) is [0, 1), which is not open in ℝ — it includes 0 as a minimum boundary point. The structural reason: f is not injective (it folds ℝ over the origin), so the image of a symmetric open interval around 0 includes 0 as an endpoint without an open neighborhood around it in the image.
More broadly, open maps tend to arise for 'quotient-like' maps that spread points out, rather than maps that fold or collapse them. Projections, quotient maps, and open linear maps are open because they don't create boundary points in the image. Maps that collapse sets (like x² folding negative numbers onto positive ones) can create boundary points in images even from open sets. The sin function is another standard example: sin(ℝ) = [−1, 1], and images of most open sets touch the endpoints ±1 without including an open neighborhood of them in [−1, 1].