Prove If K is convex, then cl(K) is convex?
prove If K is convex, then cl(K) is convex
- 9 years agoFavorite Answer
Let K be a convex subset of R^p and let x,y be in the closure of K. Then there exist sequences (x_n), (y_n) of points in K such that lim x_n = x and lim y_n = y. The line segment between x_n and y_n can be described as the set of points x_n + t * (y_n - x_n), where the parameter t ranges between 0 and 1. According to the convexity of K, each point x_n + t * (y_n - x_n) is in K. Since
lim (x_n + t * (y_n - x_n)) = lim x_n + t * (lim y_n - lim x_n) = x + t * (y-x)
and the fact that cl(K) is closed, x + t * (y-x) in in cl(K) for every t between 0 and 1. So if x,y are in cl(K), then the line segment joining x and y is also in cl(K).
- arzateLv 44 years ago
for confident HOMEWORK. 27. i visit assume that fis differentiable everywhere. hint: Use the potential fee theorem. For the counterexample, purely think of of particularly. it extremely is trivial. 28. hint: enable L be the shrink of f'. pass a techniques sufficient out so as that |f'(x)-L|<epsilon for x that a techniques out. Now %. x, y farther out and use the recommend fee theorem. 40-one. hint: think of f(x)>0 swifter or later indoors the era. enable x_0 be a place the region f is a optimal. Use the equation to tutor that f''(x_0)>0. Why is that this a contradiction? Argue further for f(x)<0. 4. it extremely is often the *definition* of convexity. what's yours?