Nuprl Lemma : real-continuity1
∀a,b:ℝ.
∀f:[a, b] ⟶ℝ
∀k:ℕ+. ∃d:{d:ℝ| r0 < d} . ∀x,y:{x:ℝ| x ∈ [a, b]} . ((|x - y| ≤ d) ⇒ (|(f x) - f y| ≤ (r1/r(k))))
supposing ∀x,y:{x:ℝ| x ∈ [a, b]} . ((x = y) ⇒ ((f x) = (f y)))
supposing a ≤ b
Proof
Definitions occuring in Statement :
rfun: I ⟶ℝ,
rccint: [l, u],
i-member: r ∈ I,
rdiv: (x/y),
rleq: x ≤ y,
rless: x < y,
rabs: |x|,
rsub: x - y,
req: x = y,
int-to-real: r(n),
real: ℝ,
nat_plus: ℕ+,
uimplies: b supposing a,
all: ∀x:A. B[x],
exists: ∃x:A. B[x],
implies: P ⇒ Q,
set: {x:A| B[x]} ,
apply: f a,
natural_number: $n
Definitions unfolded in proof :
real-fun: real-fun(f;a;b),
real-cont: real-cont(f;a;b)
Lemmas referenced :
real-continuity-ext
Rules used in proof :
hypothesis,
extract_by_obid,
introduction,
cut,
computationStep,
sqequalTransitivity,
sqequalReflexivity,
sqequalRule,
sqequalSubstitution
Latex:
\mforall{}a,b:\mBbbR{}.
\mforall{}f:[a, b] {}\mrightarrow{}\mBbbR{}
\mforall{}k:\mBbbN{}\msupplus{}
\mexists{}d:\{d:\mBbbR{}| r0 < d\} . \mforall{}x,y:\{x:\mBbbR{}| x \mmember{} [a, b]\} . ((|x - y| \mleq{} d) {}\mRightarrow{} (|(f x) - f y| \mleq{} (r1/r(k))))
supposing \mforall{}x,y:\{x:\mBbbR{}| x \mmember{} [a, b]\} . ((x = y) {}\mRightarrow{} ((f x) = (f y)))
supposing a \mleq{} b
Date html generated:
2018_05_22-PM-02_11_54
Last ObjectModification:
2018_05_21-AM-00_27_53
Theory : reals
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