Could someone please check my work

Definition 3.4.1 Let & E R and let & > 0 . A neighborhood of a (or an & - neighborhood of a ) is a set of the form N(z; E) = ly E Rx - y| 0 . A deleted neighborhood of a is a set of the form N'(x; E) = ly ER:0 0 , N (x; E) n S * 0 . The set of all accumulation points of S is denoted by S' . If a E S and a & S , then x is called an isolation point of S . An equivalent way of defining an accumulation point a of a set S would be to require that each neighborhood of x contain at least one point of S different from x . Note that an accumulation point of S may be , but does not have to be, a member of S . Prove by contradiction: If x is an accumulation point of the set S , then every neighborhood of a contains infinitely many points of S . WTS x is not an accumulation point, i.e., some & > 0 such that N(x; E) n S = 0 provides a contradiction for a being an accumulation point of S . Proof: ewollot 1 earllog ton Suppose, by contradiction, that some neighborhood of a contains only finitely many points of S. Let & E R . Choose & = min {x - xi|:1 0. Let NV be a neighborhood of x such that N n S = {x1, x2, ... In}. Then & is the smallest distance between a and any element in N n S , which means & does not have enough radius to meet any point in N n S other than itself. By definition 3.4.1 for the neighborhood of a and by the definition of set intersection, = N(x; E) ns = (x}. By definition 3.4.2 for the deleted neighborhood of xand the definition of intersection, N (x; E) nS=0. Thus, for some E > 0 , N (x; E ) nS = 0. Hence, by definition 3.4. 14 for an accumulation point of S, T is not an accumulation point of S , which is a contradiction. Therefore, by contradiction, if x is an accumulation point of the set S , then every neighborhood of x contains infinitely many points of S