Nuprl Lemma : range_sup_functionality
∀I:{I:Interval| icompact(I)} . ∀f:{x:ℝ| x ∈ I} ⟶ ℝ.
∀g:{x:ℝ| x ∈ I} ⟶ ℝ. sup{f[x] | x ∈ I} = sup{g[x] | x ∈ I} supposing ∀x:{x:ℝ| x ∈ I} . (f[x] = g[x])
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},
icompact: icompact(I),
i-member: r ∈ I,
interval: Interval,
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],
member: t ∈ T,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
prop: ℙ,
so_apply: x[s],
implies: P ⇒ Q,
uiff: uiff(P;Q),
and: P ∧ Q,
cand: A c∧ B,
exists: ∃x:A. B[x],
rev_uimplies: rev_uimplies(P;Q)
Latex:
\mforall{}I:\{I:Interval| icompact(I)\} . \mforall{}f:\{x:\mBbbR{}| x \mmember{} I\} {}\mrightarrow{} \mBbbR{}.
\mforall{}g:\{x:\mBbbR{}| x \mmember{} I\} {}\mrightarrow{} \mBbbR{}
sup\{f[x] | x \mmember{} I\} = sup\{g[x] | x \mmember{} I\} supposing \mforall{}x:\{x:\mBbbR{}| x \mmember{} I\} . (f[x] = g[x])
supposing \mforall{}x,y:\{x:\mBbbR{}| x \mmember{} I\} . ((x = y) {}\mRightarrow{} (f[x] = f[y]))
Date html generated:
2020_05_20-PM-00_19_26
Last ObjectModification:
2020_01_03-PM-03_33_35
Theory : reals
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