Nuprl Lemma : range_sup_functionality_wrt_subinterval

I:{I:Interval| icompact(I)} . ∀f:{x:ℝx ∈ I}  ⟶ ℝ.
  ∀J:{J:Interval| icompact(J)} (J ⊆ I   (sup{f[x] x ∈ J} ≤ sup{f[x] x ∈ I})) 
  supposing ∀x,y:{x:ℝx ∈ I} .  ((x y)  (f[x] f[y]))


Proof




Definitions occuring in Statement :  range_sup: sup{f[x] x ∈ I} subinterval: I ⊆  icompact: icompact(I) i-member: r ∈ I interval: Interval rleq: x ≤ y req: y real: uimplies: supposing a so_apply: x[s] all: x:A. B[x] implies:  Q set: {x:A| B[x]}  function: x:A ⟶ B[x]
Definitions unfolded in proof :  all: x:A. B[x] uimplies: supposing a member: t ∈ T implies:  Q so_lambda: λ2x.t[x] so_apply: x[s] subtype_rel: A ⊆B uall: [x:A]. B[x] subinterval: I ⊆  prop: rleq: x ≤ y rnonneg: rnonneg(x) le: A ≤ B and: P ∧ Q guard: {T}

Latex:
\mforall{}I:\{I:Interval|  icompact(I)\}  .  \mforall{}f:\{x:\mBbbR{}|  x  \mmember{}  I\}    {}\mrightarrow{}  \mBbbR{}.
    \mforall{}J:\{J:Interval|  icompact(J)\}  .  (J  \msubseteq{}  I    {}\mRightarrow{}  (sup\{f[x]  |  x  \mmember{}  J\}  \mleq{}  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_17_10
Last ObjectModification: 2020_01_03-PM-03_10_43

Theory : reals


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