Advanced Calculus
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Limit points
Definition 1.1. A sequence is a function p : N → X. We usually write a sequence p as {pn }∞
n=1 .
Definition 1.2. Let (X, d) be a metric space. We say that a sequence of points {pn }∞
n=1 converges
to p in X if
∀ > 0, ∃N ∈ N such that for n ∈ N and n ≥ N, d(pn , p) < We write pn → p, or pn → p as n → ∞, or
lim pn = p
n→∞
to indicate the convergence. We say that p is the limit of the sequence.
If S is a subset of X, we say that a point p ∈ X is a limit point of S if there exists a sequence
{pn }∞
n=1 in S that converges to p.
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Example 1.3. Let S = ∪∞
n=1 { n }. Then 0 is a limit point of S.
Lemma 1.4. Let (X, d) be a metric space and S ⊂ X. The following are equivalent.
1. p is a limit point of S.
2. For each r > 0, Br (p) ∩ S 6= ∅.
Proof. Suppose that p is a limit point of S. By definition, there is a sequence {pn }∞
n=1 in S that
converges to p. Hence for each r > 0, there is N ∈ N such that d(pn , p) < r for n > N . So
pn ∈ Br (p) ∩ S for n > N . If the second statement is true, let rn = n1 and pick pn ∈ Brn (p) ∩ S.
Then pn → p. So p is a limit point of S.
Definition 1.5. Let A be a subset of a metric space X. We write A for the set of all limit points
of A in X. A is called the closure of A in X.
Theorem 1.6. Let (X, d) be a metric space and A ⊂ X. Then A is closed in X if and only if
A = A.
Proof. Suppose that A is closed and p ∈ X is a limit point of A. If p ∈
/ A, p ∈ X − A. Since X − A
is open, there is r > 0 such that Br (p) ⊂ X − A and hence Br (p) ∩ A = ∅. This means that p is not
a limit point of A which is a contradiction. To prove the converse, suppose that all limit points of
A belong to A. Assume that A is not closed. Then X − A is not open. So there is q ∈ X − A such
that for any r > 0, Br (q) * X − A. Then Br (q) ∩ A 6= ∅. By Lemma 1.4, q is a limit point of A
and by the hypothesis, q ∈ A which is a contradiction. Hence A is closed.
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Example 1.7. Let S = ∪∞
n=1 { n }.
1. Consider S as a subset of R,
S = S ∪ {0}
2. Consider S as a subset of (0, 1] where (0, 1] is endowed with the metric inherited from R, then
S=S
Example 1.8. For S = R − Z, S = R.
Definition 1.9. In a metric space X, a subset A ⊂ X is said to be dense if
A=X
The set A is somewhere dense if there exists an open non-empty set U ⊂ X such that U ⊂ U ∩ A.
If A is not somewhere dense then it is nowhere dense.
Proposition 1.10. Let A ⊂ Z. Then A is closed in R.
Proof. Let p ∈ R − A and take n = bpc, the integer smaller than p and closest to p. Take
r = min{p − n, (n + 1) − p}
Then (p − r, p + r) ⊂ R − A. Therefore R − A is open and hence A is closed in R.
Corollary 1.11. Z is nowhere dense in R.
Proof. Assume that there exists a nonempty open subset U ⊂ R such that U ⊂ U ∩ Z. Note that
U ∩ Z ⊂ Z and hence by Proposition above, U ∩ Z is closed in R. Therefore U ⊂ U ∩ Z = U ∩ Z.
Let p ∈ U ∩ Z. Since U is open, there is 0 < r < 1 such that (p − r, p + r) ⊂ U . But p ∈ Z, p + 12 r
is not an integer, therefore (p − r, p + r) * U ∩ Z which is a contradiction. Therefore Z is nowhere
dense in R.
Example 1.12.
1. The set Q is dense in R.
2. The set N is nowhere dense in R.
3. The set {x ∈ Q|x < 0} ∪ N is somewhere dense in R.
Solution Note that for any two real numbers p < q, there is x ∈ Q such that p < x < q. For any
p ∈ R and r > 0, take x ∈ Q such that p < x < p + r. Then Br (p) ∩ Q 6= ∅ which implies that
Q = R.
Example 1.13. Is it true that A ∩ B = A ∩ B?
Solution Take A = (0, 1), B = (1, 2). Then A = [0, 1], B = [1, 2], so A ∩ B = {1} but A ∩ B = ∅.
Proposition 1.14. Let S be a subset of a metric space (X, d).
1. S = S.
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2. int(int(S)) = int(S).
3.
\
S=
C
S⊂C
C is closed
4.
[
int(S) =
U
U ⊂S
U isopen
Proof.
1. If p is a limit of S, there is a sequence {pn } in S that converges to p. Since pn is a
limit of S, there is qn ∈ S such that d(pn , qn ) < n1 . Then d(p, qn ) ≤ d(p, pn ) + d(pn , qn ) → 0
as n → ∞. Hence p is a limit of S and we have S ⊂ S. The other direction follows from the
definition.
2. For p ∈ int(S), there is r > 0 such that Br (p) ⊂ S. For each q ∈ Br (p), since Br (p) is
open, there is r0 > 0 such that Br0 (q) ⊂ Br (p) ⊂ S. Hence Br (p) ⊂ int(S). Therefore
p ∈ int(int(S)).
3. Since S is closed, the set on the right hand side is a subset of S. For p ∈ S, if p is not in the
set on right hand side, there is some C closed, S ⊂ C such that p ∈
/ C. Since X − C is open,
there is r > 0 such that Br (p) ∩ C = ∅. But S ⊂ C, Br (p) ∩ S = ∅ implies that p is not a
limit of S which is a contradiction.
S
4. Since int(S) is open and contained in S, it is a subset of
U ⊂S U . For p ∈ U for some open
U isopen
set U contained in S, by definition, p ∈ int(S).
Remark 1.15. The closure S is the smallest closed set containing S, and the interior int(S) is
the largest open set contained in S.
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