Let S be a bounded infinite set and let x = sup S. Prove : If x does not belong to S, then x belongs to S'.

Question

S' - set of all accumulation points of set S

Answer 1

Let x=sup S ∧ x∉S. Every neighborhood of x contains an open interval of the form (x-ε, x+ε) for some ε>0. Every such interval must contain at least one point from S, for otherwise x-ε would be a lower upper bound for S, contradicting the fact that x is the least upper bound. So every neighborhood of x contains a point from S, and since x∉S, this means every neighborhood of x contains a point from S other than x itself. Thus, x is an accumulation point of S. Q.E.D.

Answer 2

that is not too complicated to spell out a shape explicitly consequently. kb has published a good hyperlink to the information of the Bolzano-Weierstrass theorem, which this assertion follows from without delay. in spite of the undeniable fact that, on condition that we want in simple terms build a chain {x_n} and not a subsequence of an already-special series, we can minimize out a number of the stuff interior the completed information. bear in mind that the supremum s of a series has the valuables that x ? s for all x in S, yet on the comparable time, given any ? > 0, there is a few element x of S so as that s - ? < x ? s. that's each so often taken via fact the definition of supremum, and something of the time it follows as an instantaneous results of "s is the least top sure." the form of our series is as follows: for n = a million, 2, ..., decide for a element x_n gratifying s - a million/n < x_n ? s. on condition that a million/n > 0, the definition of supremum assures we'd try this. This inequality may be rearranged as: -a million/n < x_n - s ? 0, or equivalently, 0 ? s - x_n < a million/n. for this reason, |s - x_n| < a million/n. Now, to instruct convergence. The Archimedean materials of the reals helps us to, given any ? > 0, locate some integer N so as that ? > a million/N. For genuinely any n, |s - x_n| < a million/n. We equipped our series that way. If n > N, then a million/n < a million/N. So |s - x_n| < a million/n < a million/N < ?. hence {x_n} converges to s.