A student wants to prove f: X → Y is continuous. They verify that for every open set U in X, f(U) is open in Y. Does this prove continuity?
AYes — mapping open sets to open sets is the definition of continuity
BYes — in any topological space, forward and backward open-set conditions are equivalent
CNo — this proves f is an open map, not that f is continuous; continuity requires preimages of open sets to be open
DNo — you also need to verify f sends closed sets to closed sets
This is the most common confusion in topological continuity. Checking f(U) open for every open U is the 'open map' condition — a different and strictly stronger property. Continuity requires the preimage direction: f⁻¹(V) is open in X for every open V in Y. The constant function f(x) = 0 is continuous but maps every open set to {0}, which is not open. These are fundamentally different conditions.
Question 2 Multiple Choice
Which statement correctly characterizes when f: X → Y is continuous in the topological sense?
AFor every open set U in X, the image f(U) is open in Y
BFor every open set V in Y, the preimage f⁻¹(V) = {x ∈ X : f(x) ∈ V} is open in X
CFor every point x in X, f(x) belongs to some open subset of Y
Df maps convergent sequences in X to convergent sequences in Y
The topological definition of continuity is: f⁻¹(V) is open in X for every open V in Y. The preimage direction is essential — it asks whether the topology on X is fine enough to detect all the open structure of Y through f. Option A defines an open map. Option C is trivially true in any non-empty topological space. Option D (sequential continuity) is equivalent to topological continuity only in metric spaces, not general topological spaces.
Question 3 True / False
If f: X → Y is continuous, then f maps most open set in X to an open set in Y.
TTrue
FFalse
Answer: False
Continuity requires preimages of open sets to be open — not that images of open sets are open. The constant function f: ℝ → ℝ, f(x) = 5 is continuous (trivially, since f⁻¹(V) is either ∅ or all of ℝ for any open V), but maps the open interval (0, 1) to {5}, which is not open. The forward direction — images of open sets are open — defines an 'open map,' a strictly stronger condition that continuous functions need not satisfy.
Question 4 True / False
The topological definition of continuity (preimages of open sets are open) agrees with the ε-δ definition on metric spaces because open balls generate the metric topology.
TTrue
FFalse
Answer: True
In a metric space, the topology is generated by open balls, so every open set is a union of open balls. The ε-δ condition says: for every ε-ball around f(x), there is a δ-ball around x that maps inside it — i.e., every open ball around f(x) has an open preimage. Since open sets are unions of open balls, this generalizes to: preimages of all open sets are open. The topological definition is a true generalization: equivalent in metric spaces, but applicable without any notion of distance.
Question 5 Short Answer
Why is topological continuity defined using preimages (f⁻¹(V) open) rather than images (f(U) open)? What would go wrong if we used images instead?
Think about your answer, then reveal below.
Model answer: Using images would define an open map, not continuity — a different and stronger condition. Many continuous functions are not open maps (e.g., constant functions). The preimage direction correctly captures the intuition: continuity means that if we require the output to lie in some open region V, the set of inputs mapping into V (i.e., f⁻¹(V)) must be an open region in X. This matches the ε-δ idea: every neighborhood of the output pulls back to a neighborhood of the input.
The asymmetry is meaningful. Asking 'does f send open sets forward to open sets?' tests whether f preserves the output topology. Asking 'does f pull open sets backward to open sets?' tests whether nearby inputs produce nearby outputs — the essence of continuity. The preimage direction also makes continuity composable in the right way: if f and g are continuous, f⁻¹(g⁻¹(V)) = (g∘f)⁻¹(V) is open, so the composition is continuous. The image direction does not compose this cleanly.