Math 412/612 - hw 2 solutions
Thur, Feb 4, 2016
1. Let (X, d) be a metric space. Prove that every Cauchy sequence is bounded. Show that the converse is false.
Let (xn ) be a Cauchy sequence in X. Then there is N such that d(xm , xn ) < 1 for all m, n > N . Fix m > N . Set
M := max{1, d(xm , x1 ), . . . , d(xm , xm−1 )}. Then d(xm , xn ) ≤ M for all n ∈ N. So by the triangle inequality the
diameter of {xn : n ∈ N} is no more than 2M .
Let X := R with the usual metric. Then the sequence xn := (−1)n is is bounded but it is not Cauchy. Indeed for
every n ∈ N we hve d(xn , xn+1 ) = |(−1)n − (−1)n+1 | = 2. So if we take ε = 1 there is no N such that d(xm , xn ) < ε
for all m, n > N .
2. Let (X, d) be a metric space, let Y ⊂ X and let x ∈ X. We say that x is a boundary point of Y if every neighborhood
of x contains points in Y and points in Y c . Prove that x is a boundary point iff x ∈ Y \ int Y .
Suppose x is a boundary point of X. Then for every r > 0, there is a point y ∈ Y and z ∈ Y c such that
y, z ∈ B(x; r). Hence, x ∈ Y . Since B(x; r) 6⊂ Y for every r > 0, x 6∈ int Y . Therefore, x ∈ Y \ int Y .
Now suppose that x ∈ Y \ int Y . Then x ∈ Y but x 6∈ int Y . Let N be a neighborhood of y. Then there is r > 0
such that B(x; r) ⊂ N . Since x ∈ Y , there is y ∈ Y such that y ∈ B(x; r). But since x 6∈ int Y , B(x; r) is not
contained in Y . Thus there is z ∈ Y c such that z ∈ B(x; r). Since y, z ∈ N , x is a boundary point of X.
3. Let X be a nonempty set. Two metrics d1 and d2 are said to be equivalent if there are positive constants a, b such
that for all x, y ∈ X
ad1 (x, y) ≤ d2 (x, y) ≤ bd1 (x, y).
Let U ⊂ X. Prove that U is open with respect to d1 iff it is open with respect to d2 .
Suppose that U is open with respect to d1 and let x ∈ U . Then there is r > 0 such that if Bd1 (x, r) ⊂ U , that is,
for all y, d1 (x, y) < r implies y ∈ U . Now given y such that d2 (x, y) < ar, we have
d1 (x, y) ≤ d2 (x, y)/a < ar/a = r
and so y ∈ U . Therefore, Bd2 (x, ar) ⊂ U and so U is open with respect to d2 . The converse implication follows by
a similar argument using the second inequality.
4. Let X := [0, ∞) and let d(x, y) := |e−x − e−y | for x, y ∈ X. Given that d is a metric, prove that X is not complete.
We define a sequence (xn ) in X by xn := n. We claim that (xn ) is Cauchy. Let ε > 0, then there is N such that
0 ≤ e−x < ε for all x > N . Let m, n > N ; without loss of generality we may assume that m ≤ n. Then
d(xm , xn ) = |e−m − e−n | ≤ e−m < ε
and so (xn ) is Cauchy. Next we show that (xn ) does not converge. Let x ∈ X, then
lim d(x, xn ) = lim |e−x − e−n | = e−x > 0.
n→∞
n→∞
Therefore the sequence (xn ) does not converge to x and since x was chosen arbitrarily, (xn ) does not converge.
Therefore X is not complete.
5. (Math 612 ) Let c0 denote the space of complex sequences x = (x1 , x2 , . . . ) such that xn → 0. Show that c0 is a
closed subset of `∞ . Determine if it is complete in the induced metric.
Note that c0 ⊂ `∞ because every convergent sequence is bounded. Let y0 ∈ c0 ⊂ `∞ . Then there is a sequence
(yn ) in c0 such that yn → y0 . (Note that convergence is in the metric induced by the norm k · k∞ ). We must
show that y0 ∈ c0 , that is, y0i → 0 (where y0 = (y01 , y02 , . . . )). Let ε > 0. Then there is N1 such that
supi |y0i − yni | = ky0 − yn k∞ < ε/2 for all n > N1 . Choose m > N1 and note that |y0i − ymi | ≤ ky0 − ym k∞ < ε/2
for all i ∈ N. Since ym ∈ c0 , there is N2 such that |ymi − 0| < ε/2 for all i > N2 . Therefore, for all i > N2 we have
ε ε
|y0i − 0| ≤ |y0i − ymi | + |ymi − 0| < + = ε.
2 2
Thus y0i → 0 and consequently y0 ∈ c0 .