If F is Continuous in the Usual Metric Then Show That There Exists an Open Neighborhood
Neighbourhoods and open sets in metric spaces
Although it will not be clear for a little while, the next definition represents the first stage of the generalisation from metric to topological spaces.
An (open) -neighbourhood of a point p is the set of all points within
of it.
Definition
An open neighbourhood of a point p in a metric space (X, d) is the set V (p) = {x
X | d(x, p) <
}
Examples
- In the real line R an open neighbourhood is the open interval (p -
, p +
).
- In R 2 (with the usual metric d 2) an open neighbourhood is an "open disc" (one not containing its boundary); in R 3 it is an "open ball" etc.
- Let X be the interval [0, 1] with its usual metric. Then a 1/4 -neighbourhood of 0 is the interval [0, 1/4).
Note that this is not an open interval.
The fact that when one takes a subset of a metric space (called a subspace) the appearance of things like neighbourhoods may change is an important fact that we will need later on.
- In C[0, 1] with the metric d
one can recognise an
-neighbourhood of a point (or function) f as the set of functions whose graphs lie in an
-band around the graph of f.
Use of the idea of neighbourhood allows us the rephrase our important analytic definitions:
- A sequence (x i )
x in a metric space if every
-neighbourhood contains all but a finite number of terms of (x i ).
- A function f from a metric space X to a metric space Y is continuous at p
X if every
-neighbourhood of f(p) contains the image of some
-neighbourhood of p.
We can now use the concept of an -neighbourhood to define one of the most important ideas in a metric space.
Definition
A subset A of a metric space X is called open in X if every point of A has an -neighbourhood which lies completely in A.
Examples
- An open interval (0, 1) is an open set in R with its usual metric.
Proof
Choose< min {a, 1-a}. Then V
(a)
(0, 1).
Note, however, that there are other subsets of R which are open but which are not open intervals. For example (0, 1)
(2, 3) is an open set.
- Let X = [0, 1] with its usual metric (which it inherits from R). Then the subset [0, 1/4) is an open subset of X (but not of course of R).
- A set like {(x, y)
R 2 | x 2 + y 2 < 1} is an open subset of R 2 with its usual metric.
So also is "an open square" [a square without its boundary(0, 1)
(0, 1) ].
Proof
Any point can be in included in a "small disc" inside the square.In general, any region of R 2 given by an inequality of the form {(x, y)
R 2 | f(x, y) < 1} with f(x, y) a continuous function, is an open set.
- Any metric space is an open subset of itself. The empty set is an open subset of any metric space.
We will see later why this is an important fact.
- In a discrete metric space (in which d(x, y) = 1 for every x
y) every subset is open.
Properties of open sets
- The union (of an arbitrary number) of open sets is open.
Proof
Let xA i = A. Then s
A i for some i. Since this is open, x has an
-neighbourhood lying completely inside A i and this is also inside A.
- The intersection of finitely many open sets is open.
Proof
It is enough to show this for just two open sets A and B.
So suppose xA
B.
Then xA and so has an
1-neighbourhood lying in A. Similarly x has an
2-neighbourhood lying in B. So if
= min {
1 ,
2} this
-neighbourhood lies in both A and B and hence in A
B.
Note that the same proof will not necessarily work for the intersection of infinitely many sets since we could not be sure that min {
1 ,
2 ,
3 , ...} > 0.
For example, in R with its usual metric the intersection of open intervals:(-1/i, 1/i) = {0} which is not open.
We can now connect the concept of continuity with open sets.
- If f:X
Y is a continuous function between metric spaces and B
Y is open, then f -1(B) is an open subset of X.
If f:X
Y is a function for which f -1(B) is open in X for every open set B in Y, then f is continuous at every point of X.
Source: http://www-groups.mcs.st-andrews.ac.uk/~john/MT4522/Lectures/L8.html
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