Main derivation
S01 · implies
Before
f_n(x)=(2+3/(n+1))x+1-1/(n+1)
After
f_n\text{ continuous on }[0,1]\quad\forall n\ge0Conditions: none
Rule: polynomial · Uses: G1
Leaf basis: polynomial
Each denominator is positive and each function affine, proving continuity of every approximant.
S02 · implies
Before
0\le x\le1
After
-1\le3x-1\le2,\quad\sup_{[0,1]}|f_n-f|=2/(n+1)Conditions: none
Rule: algebra · Uses: G1
Leaf basis: algebra
The magnitude is at most two and attained at x=1, giving the exact supremum.
S03 · implies
Before
\varepsilon>0,\quad N=\lfloor6/\varepsilon\rfloor+1
After
\sup_{x\in[0,1]}|f_n(x)-f(x)|<\varepsilon/3\quad(n\ge N)Conditions: none
Rule: epsilon · Uses: S02
Leaf basis: epsilon
One N depending only on epsilon controls all points and all later terms.
S04 · implies
Before
x_0\in[0,1],\quad L_N=2+3/(N+1)>0
After
\delta=\varepsilon/(3L_N)>0
Conditions: none
Rule: algebra · Uses: G1, S03
Leaf basis: algebra
Choose an arbitrary point x_0 in the domain at which continuity is to be proved, and fix the single approximant f_N selected above.
S05 · implies
Before
x\in[0,1],\quad|x-x_0|<\delta
After
|f_N(x)-f_N(x_0)|=L_N|x-x_0|<\varepsilon/3
Conditions: none
Rule: algebra · Uses: S01, S04
Leaf basis: algebra
The affine difference supplies the exact continuity bound for the fixed approximant.
S06 · implies
Before
|f(x)-f(x_0)|\le|f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|
After
|f(x)-f(x_0)|<\varepsilon/3+\varepsilon/3+\varepsilon/3=\varepsilon
Conditions: none
Rule: algebra · Uses: S03, S05
Leaf basis: algebra
Two uniform errors and one fixed-function continuity error combine by the triangle inequality.
S07 · definition
Before
x_0\in[0,1]\text{ arbitrary},\quad\forall\varepsilon>0\ \exists\delta>0After
f\text{ continuous on }[0,1]Conditions: none
Rule: continuity_def · Uses: S04, S06, continuity_def
Leaf basis: continuity_def
Arbitrariness of x_0 establishes continuity everywhere; the same transfer in general only needs approximant continuity and uniform error.
S08 · implies
Before
u_n(x)=x^n\ (n\ge1),\quad u(x)=0\ (x<1),\ u(1)=1
After
u_n\to u\text{ pointwise but }|u_n(2^{-1/n})-u(2^{-1/n})|=1/2Conditions: none
Rule: algebra · Uses: G1
Leaf basis: algebra
Continuous powers have a pointwise limit discontinuous at one; every n has a domain point with error 1/2, so convergence is not uniform.