Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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FALSE: limxcf(g(x))=M\lim_{x \to c} f(g(x)) = M whenever limxcg=L\lim_{x \to c} g = L and limyLf=M\lim_{y \to L} f = M

Statement

False claim: let A,BRA, B \subseteq \mathbb{R}, let g:ARg : A \to \mathbb{R} with g(A)Bg(A) \subseteq B and f:BRf : B \to \mathbb{R}, let cc be a limit point of AA and LL a limit point of BB. If

limxcg(x)=LandlimyLf(y)=M,\lim_{x \to c} g(x) = L \qquad \text{and} \qquad \lim_{y \to L} f(y) = M ,

then the limit of fgf \circ g at cc exists and limxcf(g(x))=M\lim_{x \to c} f\bigl(g(x)\bigr) = M (The ε\varepsilon-δ\delta limit limxcf(x)=L\lim_{x \to c} f(x) = L of f:ARf : A \to \mathbb{R} at a limit point cc of AA).

This is the statement of Composition of limits holds under either hypothesis: ff is defined at LL with value MM, or gg avoids LL on a punctured neighbourhood of cc with both of its extra hypotheses removed, and it is false. It is refuted below by a pair in which gg is constant and ff has a removable defect at the value of that constant.

Where the naive argument breaks. The inner limit gives g(x)L<ρ|g(x) - L| < \rho for xx near cc; the outer limit gives f(y)M<ε|f(y) - M| < \varepsilon for yBy \in B with 0<yL<ρ0 < |y - L| < \rho. To combine them at y=g(x)y = g(x) one needs g(x)L>0|g(x) - L| > 0, and nothing in the hypotheses supplies that. Where g(x)=Lg(x) = L, the only information available about ff is its value f(L)f(L), and The ε\varepsilon-δ\delta limit limxcf(x)=L\lim_{x \to c} f(x) = L of f:ARf : A \to \mathbb{R} at a limit point cc of AA says nothing whatever about that value (FALSE: limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c) whenever both sides exist). The two hypotheses of Composition of limits holds under either hypothesis: ff is defined at LL with value MM, or gg avoids LL on a punctured neighbourhood of cc are exactly the two ways of closing that gap.

Facts & Assumptions

Given: The sets A:=RA := \mathbb{R} and B:=RB := \mathbb{R}; the point c:=0c := 0; the function f:RRf : \mathbb{R} \to \mathbb{R} of FALSE: limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c) whenever both sides exist, namely f(y):=0f(y) := 0 for y0y \ne 0 and f(0):=1f(0) := 1; and the constant function g:RRg : \mathbb{R} \to \mathbb{R}, g(x):=0g(x) := 0 for every xx.

[L1]

The limit condition (The ε\varepsilon-δ\delta limit limxcf(x)=L\lim_{x \to c} f(x) = L of f:ARf : A \to \mathbb{R} at a limit point cc of AA): limxch(x)=P\lim_{x \to c} h(x) = P means that for every real ε>0\varepsilon > 0 there is a real δ>0\delta > 0 such that every xx in the domain of hh with 0<xc<δ0 < |x - c| < \delta satisfies h(x)P<ε|h(x) - P| < \varepsilon.

[L3]

Absolute value: 0=0|0| = 0 (Basic properties of the absolute value).

[L4]

Order in R\mathbb{R}: trichotomy, and 0<10 < 1, so 101 \ne 0 (The multiplicative identity is positive, Ordered field).

[L5]

The function ff above satisfies f(0)=1f(0) = 1 and has limit 00 at 00: for every real ε>0\varepsilon > 0 the radius δ=1\delta = 1 works, since 0<y0<10 < |y - 0| < 1 forces y0y \ne 0 and then f(y)0=0<ε|f(y) - 0| = 0 < \varepsilon; this is the computation carried out in FALSE: limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c) whenever both sides exist.

Refutation

technique · direct
1.1

The point 00 is a limit point of R\mathbb{R}, and g(R)={0}R=Bg(\mathbb{R}) = \{0\} \subseteq \mathbb{R} = B, so fgf \circ g is a function on R\mathbb{R}.

L2
1.2

By [L5], limy0f(y)=0\lim_{y \to 0} f(y) = 0; so the outer hypothesis holds with L=0L = 0 and M=0M = 0.

L5
1.3

The reals 00 and 11 are distinct.

L4
2.1

The inner hypothesis holds with L=0L = 0: for the constant function gg and any real ε>0\varepsilon > 0, every δ>0\delta > 0 works, since g(x)0=0=0<ε|g(x) - 0| = |0| = 0 < \varepsilon for every xx. So limx0g(x)=0\lim_{x \to 0} g(x) = 0.

step 1.1L1L3
3.1

But fgf \circ g is the constant function 11: for every xRx \in \mathbb{R}, g(x)=0g(x) = 0 and hence f(g(x))=f(0)=1f(g(x)) = f(0) = 1. Therefore, by the same computation as in step 2.1, the limit of fgf \circ g at 00 exists and equals 11.

step 2.1L1L3L5
3.2

Both extra hypotheses of Composition of limits holds under either hypothesis: ff is defined at LL with value MM, or gg avoids LL on a punctured neighbourhood of cc fail for this pair: hypothesis (i) fails because L=0L = 0 lies in B=RB = \mathbb{R} while f(L)=f(0)=10=Mf(L) = f(0) = 1 \ne 0 = M; and hypothesis (ii) fails because g(x)=0=Lg(x) = 0 = L for every xx, so no punctured neighbourhood of 00 avoids the value LL.

step 2.1L5L6
4.1

So limx0g(x)=0=L\lim_{x \to 0} g(x) = 0 = L and limy0f(y)=0=M\lim_{y \to 0} f(y) = 0 = M, while limx0f(g(x))=10=M\lim_{x \to 0} f(g(x)) = 1 \ne 0 = M: the claim is false, and step 3.2 identifies exactly which hypotheses of the true theorem are missing.

step 1.2step 1.3step 3.1step 3.2

Remarks

Depends on

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 35 results over 14 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