Questions: Connected Spaces

Question 1

Consider X = ℝ² S¹, the plane with the unit circle removed (all points except those with x² + y² = 1). Is X connected?

• A. Yes — removing a curve of zero width cannot disconnect the plane
• B. Yes — you can still travel between any two points by going around the removed circle
• C. No — the open disk {x² + y² < 1} and the exterior {x² + y² > 1} form a separation into two disjoint open sets
• D. It depends on whether we use the standard or discrete topology on ℝ²

Answer: (C)

The unit circle disconnects the plane: {x² + y² < 1} and {x² + y² > 1} are both open in ℝ², both nonempty, disjoint, and their union is exactly X. This is a valid separation — so X is disconnected. Option A is the classic geometric misconception: intuitively, a curve 'has no width,' but topologically, removing it creates two pieces with no path between them that stays in X.

Question 2

The Intermediate Value Theorem — that a continuous f: [0,1] → ℝ with f(0) < c < f(1) attains the value c — is a consequence of which topological fact?

• A. The completeness of ℝ and the fact that bounded sequences have convergent subsequences
• B. Continuous images of connected spaces are connected, and connected subsets of ℝ are intervals
• C. Continuous functions on compact sets are uniformly continuous
• D. The ε-δ definition of continuity prevents functions from skipping values

Answer: (B)

The topological proof: [0,1] is connected. f is continuous, so f([0,1]) is a connected subset of ℝ. Connected subsets of ℝ are exactly the intervals. Since f(0) and f(1) are in f([0,1]) and f([0,1]) is an interval, every value between them must also be in f([0,1]). The IVT is not a special theorem — it is connectedness applied to the real line.

Question 3

In a connected topological space, the only subsets that are simultaneously open and closed (clopen) are the empty set and the entire space.

• T. True
• F. False

Answer: True

This is one of the key characterizations of connectedness. If U is a nonempty proper clopen subset, then U and its complement V = X \ U are both open, both nonempty, and disjoint with U ∪ V = X — a separation. So if X is connected, no such U can exist; the only clopens are ∅ and X. Conversely, if the only clopens are ∅ and X, any open partition would require one part to also be closed, forcing it to be ∅ or X.

Question 4

A topological space is connected if and mainly if nearly every two points in it can be joined by a continuous path.

• T. True
• F. False

Answer: False

This describes path-connectedness, which is a strictly stronger condition. Every path-connected space is connected, but not every connected space is path-connected. The classic counterexample is the topologist's sine curve: the closure of {(x, sin(1/x)) : x > 0} in ℝ². This set is connected (the two pieces cannot be separated by open sets) yet there is no continuous path from a point on the oscillating part to the point (0, 0) on the y-axis segment.

Question 5

Explain why ℝ with the standard topology is connected, while ℝ \ {0} is not.

Model answer: ℝ is connected because any attempt to split it into two disjoint nonempty open sets must leave a point uncovered: between any two open intervals, there is always a point in neither. More formally, suppose ℝ = U ∪ V with U, V open, disjoint, nonempty. Pick a ∈ U and b ∈ V; the supremum of [a, b] ∩ U must belong to both the closure of U and the closure of V, but disjoint open sets have disjoint closures — contradiction. For ℝ \ {0}: write it as (−∞, 0) ∪ (0, ∞). Both pieces are open in ℝ \ {0}, nonempty, and disjoint — a valid separation.

Removing a single point from ℝ creates a gap that cannot be bridged by any path remaining in ℝ \ {0}. Topologically, the removed point was the only 'link' between the two halves, and its absence makes the separation exact.