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Exercise 1185Topology › Topology of ℝL1Added 3 Nov 2026

Connected subsets of $\mathbb{R}$

#connectedness#intervals#IVT

Statement

Show that the connected subsets of R\mathbb{R} are exactly the intervals.

Hint

If AA has a gap, split A∩]−∞,c[A\cap]-\infty,c[ and A∩]c,+∞[A\cap]c,+\infty[. Conversely, if AA is an interval, any separation would contradict the IVT for the identity.

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Solution

Interval connected. If AA is an interval and A=U∪VA=U\cup V with U,VU,V disjoint nonempty open in AA, take u∈Uu\in U, v∈Vv\in V, WLOG u<vu<v. Since AA is an interval, [u,v]⊆A[u,v]\subseteq A. The sets U∩[u,v]U\cap[u,v] and V∩[u,v]V\cap[u,v] give a separation of [u,v][u,v], contradiction with connectedness of intervals.

Non-interval disconnected. If a<b<ca<b<c with a,c∈Aa,c\in A, b∉Ab\notin A, then A∩]−∞,b[A\cap]-\infty,b[ and A∩]b,+∞[A\cap]b,+\infty[ separate AA.

Conclusion. Connected subsets of R\mathbb{R} = intervals.