Nuprl Lemma : real-fun-bounded
∀a:ℝ. ∀b:{b:ℝ| a ≤ b} . ∀f:[a, b] ⟶ℝ.
∃bnd:ℝ. ((r0 ≤ bnd) ∧ (∀x:ℝ. ((x ∈ [a, b]) ⇒ (|f x| ≤ bnd))))
supposing ∀x,y:{x:ℝ| x ∈ [a, b]} . ((x = y) ⇒ (f[x] = f[y]))
Proof
Definitions occuring in Statement :
rfun: I ⟶ℝ,
rccint: [l, u],
i-member: r ∈ I,
rleq: x ≤ y,
rabs: |x|,
req: x = y,
int-to-real: r(n),
real: ℝ,
uimplies: b supposing a,
so_apply: x[s],
all: ∀x:A. B[x],
exists: ∃x:A. B[x],
implies: P ⇒ Q,
and: P ∧ Q,
set: {x:A| B[x]} ,
apply: f a,
natural_number: $n
Definitions unfolded in proof :
all: ∀x:A. B[x],
uimplies: b supposing a,
member: t ∈ T,
implies: P ⇒ Q,
uall: ∀[x:A]. B[x],
so_apply: x[s],
rfun: I ⟶ℝ,
exists: ∃x:A. B[x],
iff: P ⇐⇒ Q,
and: P ∧ Q,
sq_stable: SqStable(P),
squash: ↓T,
prop: ℙ,
so_lambda: λ2x.t[x],
guard: {T},
cand: A c∧ B
Latex:
\mforall{}a:\mBbbR{}. \mforall{}b:\{b:\mBbbR{}| a \mleq{} b\} . \mforall{}f:[a, b] {}\mrightarrow{}\mBbbR{}.
\mexists{}bnd:\mBbbR{}. ((r0 \mleq{} bnd) \mwedge{} (\mforall{}x:\mBbbR{}. ((x \mmember{} [a, b]) {}\mRightarrow{} (|f x| \mleq{} bnd))))
supposing \mforall{}x,y:\{x:\mBbbR{}| x \mmember{} [a, b]\} . ((x = y) {}\mRightarrow{} (f[x] = f[y]))
Date html generated:
2020_05_20-PM-00_22_29
Last ObjectModification:
2020_01_06-PM-02_44_40
Theory : reals
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