Nuprl Lemma : range_sup-property
∀I:{I:Interval| icompact(I)} . ∀f:{x:ℝ| x ∈ I} ⟶ ℝ.
sup(f[x](x∈I)) = sup{f[x] | x ∈ I} supposing ∀x,y:{x:ℝ| x ∈ I} . ((x = y) ⇒ (f[x] = f[y]))
Proof
Definitions occuring in Statement :
range_sup: sup{f[x] | x ∈ I},
rrange: f[x](x∈I),
icompact: icompact(I),
i-member: r ∈ I,
interval: Interval,
sup: sup(A) = b,
req: x = y,
real: ℝ,
uimplies: b supposing a,
so_apply: x[s],
all: ∀x:A. B[x],
implies: P ⇒ Q,
set: {x:A| B[x]} ,
function: x:A ⟶ B[x]
Definitions unfolded in proof :
all: ∀x:A. B[x],
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
member: t ∈ T,
sq_stable: SqStable(P),
implies: P ⇒ Q,
squash: ↓T,
subinterval: I ⊆ J ,
prop: ℙ,
ifun: ifun(f;I),
real-fun: real-fun(f;a;b),
subtype_rel: A ⊆r B,
so_lambda: λ2x.t[x],
and: P ∧ Q,
so_apply: x[s],
cand: A c∧ B,
range_sup: sup{f[x] | x ∈ I},
label: ...$L... t,
rfun: I ⟶ℝ,
exists: ∃x:A. B[x],
r-ap: f(x),
pi1: fst(t),
icompact: icompact(I)
Latex:
\mforall{}I:\{I:Interval| icompact(I)\} . \mforall{}f:\{x:\mBbbR{}| x \mmember{} I\} {}\mrightarrow{} \mBbbR{}.
sup(f[x](x\mmember{}I)) = sup\{f[x] | x \mmember{} I\} supposing \mforall{}x,y:\{x:\mBbbR{}| x \mmember{} I\} . ((x = y) {}\mRightarrow{} (f[x] = f[y]))
Date html generated:
2020_05_20-PM-00_15_40
Last ObjectModification:
2020_01_03-PM-02_35_05
Theory : reals
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