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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

**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.

- The union (of an arbitrary number) of open sets is open.

**Proof**

Let*x**A*_{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*x**A**B*.

Then*x**A*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*.**Proof**

Note that even if*f*is not a one-one-correspondance (and does not have an inverse function*f*^{-1}) the set*f*^{-1}(*B*) ={*x**X*|*f*(*x*)*B*} still exists.

Take*x**f*^{-1}(*B*). Then*f*(*x*) =*y**B*.

Since*B*is open the point*y*has an -neighbourhood*B*.

Then, from the definition of continuity this -neihbourhood contains the image of some -neighbourhood*V*of*x*. Since*f*(*V*)*B*we have*V**f*^{-1}(*B*) and so*x*has this neighbourhood*f*^{-1}(*B*).

Hence*f*^{-1}(*B*) is open in*X*.

The converse of this result also holds: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*.**Proof**

To show that*f*is continuous at*x**X*, take*B*to be an -neighbourhood of*y**Y*.

Then*f*^{-1}(*B*) is open in*X*and so*x*has a -neighbourhood in*f*^{-1}(*B*). This -neighbourhood is mapped inside the -neighbourhood of*f(x)*and so*f*is continuous at*x*.

We will see later that the above properties of open sets in metric spaces are the basis for the generalisation from metric to topological spaces.

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