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To be technically complete, therefore, I could have written this: "You can't read a Kindle book on a Nook or Sony Reader, or a Nook book on a Sony Reader or Kindle, or a Sony book on an iPad, Kindle or Nook, or an iBooks book on a Nook, Kindle or Sony Reader.
Since E is complete, therefore { x n } is convergent.
Since ( X, d ) is complete, therefore, g ( x n ) and g ( y n ) are convergent in ( X, d ).
Since f ( E ) = f ( X ) is complete, therefore by Corollary 25 there exists z ∈ X such that g ( f z ) = f z, which implies T z = f z.
Since ( X, d ) is tvs-cone complete, therefore { x n } is tvs-cone convergent in X and cone- lim n → ∞ x n = v.
Then the metric induced by p is given by d ( x, y ) = 3 | x − y | for all x, y ∈ X and ( X, d ) is not complete, therefore ( X, p ) is not complete.
Similar(43)
Since M is ω-complete, therefore ({x_{n}}) ω-converges to some point (xin M).
Since ((X,F,T)) is G-complete, therefore (x_{n} to u), as (n toinfty), for some (u in X).
Since ((X,M,ast)) is G-complete, therefore ({x_{n}}) converges to (x^) for some (x^ in X).
Since ((X,F,Delta)) is G-complete, therefore (lim_{nrightarrowinfty}Ax_{n}=u, lim_{nrightarrowinfty}Ay_{n}=v, ldots, lim_{nrightarrowinfty}Az_{n}=w) for some (u,v,ldots,w in X).
Since C is w-complete, therefore { T n ( x ) } w-converges to some point x 0 ∈ C. Next let us show that x 0 is a fixed point of T. Indeed, we have w 2 ( x 0, T ( x 0 ) ) ≤ w 1 ( x 0, T n ( x ) ) + w 1 ( T ( x 0 ), T n ( x ) ), which implies w 2 ( x 0, T ( x 0 ) ) ≤ w 1 ( x 0, T n ( x ) ) + k w 1 ( x 0, T n − 1 ( x ) ). for any n ≥ 1.
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