Questions: Connected Sets

Question 1

Consider S = ℝ \ {0} = (-∞, 0) ∪ (0, ∞). Is S connected?

• A. Yes — removing a single point cannot disconnect a set
• B. No — S can be written as the union of two non-empty, disjoint sets that are open in the subspace topology on S
• C. Yes — S is an open set, and open sets in ℝ are connected
• D. No — S is disconnected because it is not closed

Answer: (B)

The key is the subspace topology. In the subspace topology on S, the sets (-∞, 0) and (0, ∞) are both open (each is the intersection of S with an open interval in ℝ), non-empty, disjoint, and their union is S. This is exactly the definition of disconnectedness. Option A mistakes removing a single point for a 'small' change — in topology, even one removed point can disconnect a set. Options C and D confuse connectedness with being open or closed, which are separate properties.

Question 2

Why does the Intermediate Value Theorem follow directly from the fact that continuous images of connected sets are connected?

• A. Because continuous functions map open sets to open sets, preserving the interval structure
• B. Because if f is continuous on [a,b], then f([a,b]) is a connected subset of ℝ — which means it is an interval — so f must take every value between f(a) and f(b)
• C. Because differentiable functions cannot skip values, and continuous functions are almost everywhere differentiable
• D. Because [a,b] is both open and closed, forcing f([a,b]) to be an interval

Answer: (B)

The IVT proof via connectedness is elegant precisely because it reduces to two facts: (1) [a,b] is connected (it is an interval, and intervals are exactly the connected subsets of ℝ), and (2) continuous images of connected sets are connected. Together they force f([a,b]) to be a connected subset of ℝ — hence an interval. If f(a) and f(b) are both in f([a,b]), then every value between them must be too (since an interval contains all points between any two of its elements). This is the IVT, with no ε-δ case analysis required.

Question 3

In ℝ, a set is connected if and only if it is an interval (where single points, rays, and the empty set count as degenerate intervals).

• T. True
• F. False

Answer: True

This is the complete characterization of connected subsets of ℝ. An interval cannot be split into two disjoint non-empty open parts because any gap would create such a split. Conversely, any non-interval subset of ℝ has a gap — a point not in the set between two points that are — and this gap immediately provides the separation required by the definition of disconnectedness. The equivalence 'connected ⟺ interval' is specific to ℝ; in higher dimensions, connected does not imply path-connected or convex.

Question 4

A set S ⊆ ℝ is disconnected mainly if it can be written as a union of two disjoint non-empty closed sets (in the subspace topology).

• T. True
• F. False

Answer: False

The standard definition of disconnectedness uses open sets: S is disconnected if S = U ∪ V with U, V non-empty, disjoint, and open in the subspace topology. However, in the subspace topology, U and V would each also be closed (since each is the complement of the other within S). So for subsets of ℝ, the open-set and closed-set formulations are actually equivalent — but the statement as posed is misleading because it suggests closed sets are the fundamental requirement, when the definition is given in terms of open sets. More importantly, not all disconnected sets can be split by globally closed sets in ℝ; the subspace topology is what matters.

Question 5

Using the definition precisely, explain why the set {0, 1} ⊂ ℝ (with the subspace topology inherited from ℝ) is disconnected.

Model answer: Let U = {0} and V = {1}. In the subspace topology on {0,1}, a set is open if it equals the intersection of {0,1} with some open set in ℝ. Since {0} = {0,1} ∩ (-1/2, 1/2) and {1} = {0,1} ∩ (1/2, 3/2), both {0} and {1} are open in the subspace topology. They are also non-empty, disjoint, and their union is {0,1}. This is exactly the definition of disconnectedness: a separation of {0,1} into two disjoint non-empty open sets.

The subtlety is that 'open in the subspace topology' is not the same as 'open in ℝ.' The set {0} is not open in ℝ, but it is open in the subspace topology on {0,1} because it is the restriction of an open interval to the subspace. This is why the definition specifies 'open in the subspace topology' — it makes disconnectedness a property of the set itself, independent of how it sits in a larger space.