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MATH 3402
Metric Space Topology
Open sets
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That is, for every x ∈ S; if y ∈ X and d(y, x) < x , then y ∈ S
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It is often referred to as an ”open -neighbourhood”
or ”open -ball”
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If x ∈ N , then d(x, a) = r <
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X is open in (X, d)
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If the metric d is the discrete metric, then every subset S of X is open in (X, d)
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The union of any number of open sets in (X, d) is open in (X, d)
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Since Sa is open, there is N (x, ) ⊂ Sa
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α

The intersection of a finite number of open sets in (X, d) is open in (X, d)
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Since Si is open, there is an i > 0 such that N (x, i ) ⊂ Si
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Then > 0, and
N (x, ) ⊂ N (x, i ) ⊂ Si for each i
Therefore

n

N (x, ) ⊂

Si
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For example, the sets
Sn = {|z − z0 | < R +
are open in (C, |
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Closed sets
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neighbourhood of

A subset S of X is closed in (X, d) if S contains all its accumulation points
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The complement is taken with respect to the set X
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Therefore there is some > 0 such that N (x, ) contains no points of S
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Conversely, if \S is open, then if x ∈ \S, there is an epsilon neighbourhood of x
such that N (x, ) ⊂ \S
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From de Morgan’s laws it follows that the intersection of any number of closed
sets is closed, and the union of any finite number of closed sets is closed
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The sets
1
Sn = {|z − z0 | ≤ R − }
n
are closed in (C, |
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X is closed in (X, d)
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If d is the discrete metric, every subset of (X, d) is closed
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A function f from (X, dX ) to (Y, dY ) is continuous at a ∈ X, if, given any > 0,
there exists a δ( , a) such that
dY (f (x), f (a)) <

∀ x ∈ X ; dX (x, a) < δ
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A function f from (X, dX ) to (Y, dY ) is continuous at a ∈ X if and only if for
any sequence {an } in (X, dX ) which converges to a, the sequence {f (an )} converges
in (Y, dY ) to f (a)
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Theorem
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Proof
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If a ∈ f −1 (U ), then f (a) ∈ U
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But f is continuous at a
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Therefore N (a, δ) ⊂ f −1 (U ), and f −1 (U ) is open
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For any a ∈ X, N (f (a), ) is open in Y
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Therefore for some δ > 0, N (a, δ) ⊂ f −1 (N (f (a), )) and hence f (N (a, δ)) ⊂
N (f (a), )
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If therefore two metrics on X give rise to precisely the same open sets in X, then
any function continuous with respect to one metric will be continuous with respect
to the other
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Two metrics d1 and d2 on a space X are topologically equivalent
if and only if a subset U of X which is open in (X, d1 ) is open in (X, d2 )
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b) f : Y → X is continuous from (Y, dY ) to (X, d1 ) if and only if it is continuous
from (Y, dY ) to (X, d2 )
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a): Suppose that d1 and d2 are topologically equivalent, and that f is continuous
from (X, d1 ) to (Y, dY )
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Since d1 and d2 are topologically equivalent, f −1 (U ) is also open in (X, d2 ), so
that f is continuous from (X, d2 ) to (Y, dY )
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Since this does not depend on f or Y , we are free to choose the image space as
(X, d1 ) and the function from X to X as the identity function
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b): Follows in similar fashion
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If there are strictly positive constants c1 and c2 such that
c1 d1 (x, y) ≤ d2 (x, y) ≤ c2 d1 (x, y)
for all x, y ∈ X, then d1 and d2 are topologically equivalent metrics on X
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Let U be open in (X, d1 )
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But then
{x ∈ X; d2 (x, a) < c1 } ⊂ {x ∈ X; d1 (x, a) < } ⊂ U
and U is open in (X, d2 )
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This criterion is sufficient but not necessary
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On the other hand
{d1 (x, a) < } ⊂ {d2 (x, a) < }
so that if U is open in (R, d2 ) it is open in (R, d1 ), while if |x − a| <
d1 (x, a) = (1 + |x − a|)d2 (x, a)
≤ (1 + )d2 (x, a)
Let 1 = /(1 + )
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1}

⊂ {d1 (x, a) < }, so that if U is open in (R, d1 ) it is open

For example, in R2 ,
max(|x1 − x2 |, |y1 − y2 |)
≤ |x1 − x2 | + |y1 − y2 |
≤ 2 max(|x1 − x2 |, |y1 − y2 |)
so that the taxi-cab and sup metrics are equivalent
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