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CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)
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A real-valued continuous map on a connected space has order-convex image, so it takes every value between any two of its values

Statement

Let XX be a connected topological space (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets) and let f:XRf : X \to \mathbb{R} be continuous (Continuity of a map of topological spaces at a point and globally), R\mathbb{R} carrying its usual topology (The absolute value makes R\mathbb{R} a metric space: d(x,y)=xyd(x,y) = |x-y| is a metric, its open balls are the intervals (xr,x+r)(x-r, x+r), and it is unbounded, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). Then:

  1. f[X]f[X] is an order-convex subset of R\mathbb{R} (Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length).
  2. Intermediate values are attained. If p,qXp, q \in X and cRc \in \mathbb{R} satisfies f(p)cf(q)f(p) \le c \le f(q), then there is xXx \in X with f(x)=cf(x) = c.

Claim 2 is the intermediate value theorem with no hypothesis on XX beyond connectedness: no order, no metric, no interval. The classical statement for a continuous f:[a,b]Rf : [a,b] \to \mathbb{R} is the special case X=[a,b]X = [a,b], that subspace being connected by The connected subspaces of R\mathbb{R} with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in R\mathbb{R}".

Facts & Assumptions

Proof

technique · direct
1.1

f[X]f[X] is a connected subset of R\mathbb{R}, by [A1] applied to the connected space XX and the continuous map ff.

A1given
2.1

Hence f[X]f[X] is order-convex by [A2]; this is claim 1.

step 1.1A2
3.1

For claim 2, let p,qXp, q \in X and cRc \in \mathbb{R} with f(p)cf(q)f(p) \le c \le f(q). Both f(p)f(p) and f(q)f(q) lie in f[X]f[X], so cf[X]c \in f[X] by step 2.1, which says precisely that c=f(x)c = f(x) for some xXx \in X.

step 2.1

Remarks

  • Why f(p)cf(q)f(p) \le c \le f(q) and not f(p)<c<f(q)f(p) < c < f(q). Order-convexity is stated with non-strict inequalities, so the endpoints are included and the statement covers c=f(p)c = f(p) and c=f(q)c = f(q) without a separate clause. No assumption f(p)f(q)f(p) \le f(q) is needed either: if f(q)cf(p)f(q) \le c \le f(p) the same argument applies with the two points exchanged.

  • What is not claimed. Nothing here says that f[X]f[X] is an interval in the sense of one of the nine written forms of Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length; that classification of the order-convex sets is recorded there as unproved. Nor does the corollary say anything about how many xx satisfy f(x)=cf(x) = c, or that such an xx can be found by any procedure. It is an existence statement obtained by transporting connectedness, and the witness is never exhibited.

  • The hypothesis on XX is exactly connectedness. If XX is disconnected the conclusion fails at once: a separation (U,V)(U,V) of XX gives a continuous ff equal to 00 on UU and 11 on VV whose image is {0,1}\{0,1\}, which omits every value strictly between.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 93 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources