Consider the constant function f: ℝ → ℝ defined by f(x) = 0. Which of the following correctly applies the topological definition of continuity?
Af is not continuous because the image f(ℝ) = {0} is not open in ℝ
Bf is continuous because f⁻¹(V) is open in ℝ for every open set V in ℝ
Cf is continuous because f sends every open set in ℝ to an open set in ℝ
Df is not continuous because no ε-δ bound exists that works globally
For any open set V in ℝ: if 0 ∈ V then f⁻¹(V) = ℝ, which is open; if 0 ∉ V then f⁻¹(V) = ∅, which is also open. So f is continuous. The image {0} being non-open is irrelevant — continuity requires preimages of open sets to be open, not images. Option C describes an 'open map,' a separate and stronger condition.
Question 2 Multiple Choice
The topological definition of continuity uses preimages of open sets rather than epsilon-delta conditions primarily because:
APreimages are easier to compute than epsilon-delta bounds in concrete analysis
BThe epsilon-delta definition is incorrect even for functions on ℝ
CThe preimage condition can be stated on any topological space, including those with no notion of distance
DPreimages of open sets are always open, making continuity automatic in any space
The epsilon-delta definition is built around absolute values — a distance measure — and cannot be stated for a space with no metric. The preimage definition uses only the open sets (the topology), which can be specified on any set regardless of distance. This unifies continuity across real analysis, abstract algebra, geometry, and functional analysis under a single definition.
Question 3 True / False
If f: X → Y is continuous and C is a closed set in Y, then f⁻¹(C) is closed in X.
TTrue
FFalse
Answer: True
Since C is closed, Y \ C is open. Continuity means f⁻¹(Y \ C) is open in X. But f⁻¹(Y \ C) = X \ f⁻¹(C), so X \ f⁻¹(C) is open, which means f⁻¹(C) is closed. The open-set and closed-set characterizations of continuity are fully equivalent — preimages commute with complements.
Question 4 True / False
If f: X → Y is a continuous function between topological spaces, then the image of most open set in X is open in Y.
TTrue
FFalse
Answer: False
This is the most common confusion about topological continuity. The constant function f(x) = c maps every open set to the single point {c}, which is not open. Continuity only guarantees that preimages — not images — of open sets are open. A function that sends open sets to open sets is called an 'open map,' which is a distinct property that continuous functions need not have.
Question 5 Short Answer
Why does the topological definition of continuity use preimages rather than forward images? What would go wrong if we instead defined continuity as 'the image of every open set is open'?
Think about your answer, then reveal below.
Model answer: Continuity captures the idea that outputs near f(x) come from inputs near x — i.e., f doesn't 'tear' or 'jump.' The preimage f⁻¹(V) asks which inputs lead to outputs in the neighborhood V; requiring this to be open ensures inputs producing outputs near f(x) must themselves be near x. An image-based definition would fail for constant functions, which are continuous in every reasonable sense but send all open sets to single (non-open) points. The image direction defines 'open maps,' a genuinely different property.
The arrow reversal reflects the asymmetry of continuity: you specify an output tolerance and ask how tightly the input must be controlled. That is the preimage direction.