6.1 CONCLUSIONS INVOLVING V, FOLLOWED BY 3 2M- (a) Prove that 3 is an interior point of [O, 1).
(b) Prove that 1 is not an interior point of [0, 11.
(c) Prove that the set N of all positive integers has no interior points. Is N open
in R.
(d) Prove that any interval of the form (a, b), where a # b, is an open subset of R.
(e) Prove that R is an open subset of itself.
(f) Prove that if S,, S,, and S, are open subsets of R, then both S, n S, n S3
and S, u S, u S, are open subsets of R.
(g) Use an induction argument to prove that the union and intersection of any
finite collection of open subsets of R is an open subset of R.
(h) Generalize one of the arguments in Example 9 to show that if (Sili = 1,2,3,.. .)
is a family of open subsets of R, indexed by N, then Uim_, Si is open in R. Does
the other argument in Example 9 generalize to such an infinite collection, that
is, can we conclude that (-)El Si is necessarily open in R? - Let S be a subset of R. A real number a, which may or may not lie in S, is
said to be a point of accumulation of S if and only if every neighborhood N(a; 6)
of a contains points of S other than a itself. This definition may be represented
(V6 > O)(N1(a; 6) n S # a), where N1(a; 6) = N(a; 6) - {a} is called the deleted 6
neighborhood of a. S is said to be a closed subset of R if and only if S contains
all its points of accumulation.
(a) Prove that 1 is a point of accumulation of (0, 1). (Thus a point of accumu-
lation of a set need not lie in the set.)
*(b) Write the logical negation of the definition of point of accumulation and use
it to prove that 2 is not a point of accumulation of (0, 1) u (2). (Thus a point
in a set need not be a point of accumulation of that set.)
(c) Prove that, for any p > 0, -p is not a point of accumulation of [0, 11.
(d) Prove that 0 is a point of accumulation of the set S = (1, f, 4,.. .}. Is this set
closed? (Note: You may assume the Archimedean property, introduced following
Example 2, Article 4.2.)
(e) Prove that the set N = (1,2,3,.. .} of all positive integers has no accumu-
lation points. Is this set closed in R?
(f) Prove that any interval of the form [a, b], where a # b, is closed in R.
(g) Prove that R is a closed subset of itself.
4--- I--- ..--
- (Continuation of Exercises 8 and 9) (a) Give an example of a subset of R that
is neither open nor closed.
(b) Give an example of a subset of R that is both open and closed.
(c) Prove that if a subset S of R is closed in R, then its complement S' is open in
R. (The converse is true as well; it will be considered in Exercise 1 l(a), Article 6.2.) - Throughout our discussion of limit in Article 4.3, whenever we considered
lim,,, f(x), we assumed f to be a function defined in an open interval containing
a. In the terminology of Exercise 9 this assumption implies that a is a point of
accumulation of the domain off. Prove that if a is not a point of accumulation
of the domain off [recall Exercise 9(b)], then L = lim,,, f(x) is true for any value
of L (thus rendering the definition of limit worthless in that circumstance).