Question

in a metric space a set is open if and only if it is a neighborhood of each of its point. show it​

Answer 1

Answer:

An (open) epsilon-neighbourhood of a point p is the set of all points within epsilon of it.

Definition

An open neighbourhood of a point p in a metric space (X, d) is the set Vepsilon(p) = {x belongs X | d(x, p) < epsilon}

Examples

In the real line R an open neighbourhood is the open interval (p - epsilon, p + epsilon).

In R2 (with the usual metric d2) an open neighbourhood is an "open disc" (one not containing its boundary); in R3 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 dinfinity one can recognise an epsilon-neighbourhood of a point (or function) f as the set of functions whose graphs lie in an epsilon-band around the graph of f.

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

the real line R an open neighbourhood is the open interval (p - epsilon, p + epsilon).

In R2 (with the usual metric d2) 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 dinfinity one can recognise an epsilon-neighbourhood of a point (or function) f as the set of functions whose graphs lie in an epsilon-band around the graph of f.

Use of the idea of neighbourhood allows us the rephrase our important analytic definitions:

A sequence (xi) rarrow x in a metric space if every epsilon-neighbourhood contains all but a finite number of terms of (xi).

A function f from a metric space X to a metric space Y is continuous at p belongs X if every epsilon-neighbourhood of f(p) contains the image of some delta-neighbourhood of p.

We can now use the concept of an epsilon-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 epsilon-neighbourhood which lies completely in A.

Examples

An open interval (0, 1) is an open set in R with its usual metric.

Proof

Choose epsilon < min {a, 1-a}. Then Vepsilon(a) subset (0, 1).

Note, however, that there are other subsets of R which are open but which are not open intervals. For example (0, 1) union (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) belongs R2 | x2 + y2 < 1} is an open subset of R2 with its usual metric.

So also is "an open square" [a square without its boundary iso (0, 1) cross (0, 1) ].

Proof

Any point can be in included in a "small disc" inside the square.

In general, any region of R2 given by an inequality of the form {(x, y) belongs R2 | 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 note q y) every subset is open.