F Continuous if and Only if Pre image of Open Set is Open

In all but the last section of this wiki, the setting will be a general metric space $(X,d).$ Those readers who are not completely comfortable with abstract metric spaces may think of $X$ as being ${\mathbb R}^n,$ where $n=2$ or $3$ for concreteness, and the distance function $d(x,y)$ as being the standard Euclidean distance between two points.

An open set in a metric space $(X,d)$ is a subset $U$ of $X$ with the following property: for any $x \in U,$ there is a real number $\epsilon > 0$ such that any point in $X$ that is a distance $<\epsilon$ from $x$ is also contained in $U.$

For any point $x \in X,$ define $B(x,\epsilon)$ to be the open ball of radius $\epsilon,$ which is the set of all points in $X$ that are within a distance $\epsilon$ from $x.$ (For instance, if $X = {\mathbb R},$ then $B(x,\epsilon)$ is the open interval $(x-\epsilon,x+\epsilon).$ If $X={\mathbb R}^2,$ $B(x,\epsilon)$ is the open disk centered at $x$ with radius $\epsilon.)$ Then a set $U$ is open if and only if for each point $x\in U,$ there is an $\epsilon>0$ such that $B(x,\epsilon)$ is completely contained in $U.$

Some references use $B_{\epsilon}(x)$ instead of $B(x,\epsilon).$ ^[1]

So the intuition is that an open set is a set for which any point in the set has a small "halo" around it that is completely contained in the set. The idea is that this halo fails to exist precisely when the point lies on the boundary of the set, so the condition that $U$ is open is the same as saying that it doesn't contain any of its boundary points. With the correct definition of boundary, this intuition becomes a theorem.

The boundary of a set $S$ inside a metric space $X$ is the set of points $s$ such that for any $\epsilon>0,$ $B(s,\epsilon)$ contains at least one point in $S$ and at least one point not in $S.$

A subset $U$ of a metric space is open if and only if it does not contain any of its boundary points.

It is clear that an open set $U$ cannot contain any of its boundary points since the halo condition would not apply to those points. On the other hand, if a set $U$ doesn't contain any of its boundary points, that is enough to show that it is open: for every point $x\in U,$ since $x$ is not a boundary point, that implies that there is some ball around $x$ that is either contained in $U$ or contained in the complement of $U.$ But every ball around $x$ contains at least one point in $U,$ namely $x$ itself, so it must be the former, and $x$ has a halo inside $U.$ $_\square$

• Trivial open sets: The empty set and the entire set $X$ are both open. This is a straightforward consequence of the definition.

• Union and intersection: The union of an arbitrary collection of open sets is open. The intersection of finitely many open sets is open.

To see the first statement, consider the halo around a point in the union. Since any $x$ in the union is in one of the open sets $U,$ it has a $B(x,\epsilon)$ around it contained in $U,$ so that ball is contained in the union as well. $_\square$

The second statement is proved in the below exercise.

The intersection of infinitely many sets is not necessarily defined The proof works; the statement is true $B$ might not be contained inside $U$ $U$ might be empty $B$ might not be a ball around $x$

Let $U_{\alpha}$ $(\alpha \in A)$ be a collection of open sets in ${\mathbb R}^2.$ If $A$ is finite, then the intersection $U = \bigcap\limits_\alpha U_{\alpha}$ is also an open set. Here is a proof:

Suppose $x\in U.$ For each $\alpha \in A,$ let $B_{\alpha}$ be a ball of some positive radius around $x$ which is contained entirely inside $U_{\alpha}.$ Then the intersection of the $B_{\alpha}$ is a ball $B$ around $x$ which is contained entirely inside the intersection, so the intersection is open.

$($ Here a ball around $x$ is a set $B(x,r)$ ( $r$ a positive real number) consisting of all points $y$ such that $|x-y|<r.$ In ${\mathbb R}^2$ it is an open disk centered at $x$ of radius $r.)$

Where does this proof go wrong when $A$ is infinite?

• Basic open sets: Every open set is the union of open balls $B(x,\epsilon).$

To see this, let $U$ be an open set and, for each $x\in U,$ let $B(x,\epsilon)$ be the halo around $x.$ Then each $B(x,\epsilon)$ is contained in $U,$ so their union is; but their union must be all of $U$ since every point $x\in U$ is contained in (at least) one of them.

This notion of building up open sets by taking unions of certain types of open sets generalizes to abstract topology, where the building blocks are called basic open sets, or a base.

The definition of continuous functions, which includes the epsilon-delta definition of a limit, can be restated in terms of open sets. This reformulation turns out to be the way to generalize the concept of continuity to abstract topological spaces.

Recall that a function $f \colon {\mathbb R}^n \to {\mathbb R}^m$ is said to be continuous if $\lim\limits_{x\to a} f(x) = f(a).$

A function $f \colon {\mathbb R}^n \to {\mathbb R}^m$ is continuous if and only if the inverse image of any open set is open. That is, if $V$ is an open subset of $Y,$ then $f^{-1}(V)$ is an open subset of $X.$

Suppose $f$ is continuous, $V \subseteq {\mathbb R}^m$ is open, and $a \in f^{-1}(V).$ Then $f(a) \in V,$ so there is an open ball $B\big(f(a),\epsilon\big) \subseteq V,$ for some $\epsilon.$ Now since $\lim\limits_{x\to a} f(x) = f(a),$ there must exist some $\delta > 0$ such that whenever $|x-a|<\delta,$ $|f(x)-f(a)|<\epsilon.$ That is, for all $x \in B(a,\delta),$ $f(x)$ lies in $B\big(f(a),\epsilon\big).$ But this ball is contained in $V,$ so for all $x \in B(a,\delta),$ $f(x) \in V.$ So $B(a,\delta) \subseteq f^{-1}(V).$ This shows that $f^{-1}(V)$ is open, since we have found a ball around any point $a \in f^{-1}(V)$ which is contained in $f^{-1}(V).$

The proof of the opposite ("if") direction is similar. $_\square$

Note that the image of an open set under a continuous function is not necessarily open. For instance, $f \colon {\mathbb R} \to {\mathbb R}$ defined by $f(x)=x^2$ satisfies $f\big((-1,1)\big) = [0,1).$

The theorem above motivates the general definition of topological continuity: a continuous function between two metric spaces (or topological spaces) is defined to be a function with the property that the inverse image of an open set is open.

• The interior of a set $X$ is defined to be the largest open subset of $X.$ It equals the union of every open subset of $X.$ The interior of $X$ is the set of points in $X$ which are not boundary points of $X.$

• A connected set is defined to be a set which is not the disjoint union of two nonempty open sets.

• A compact subset of ${\mathbb R}^n$ is a subset $X$ with the property that every covering of $X$ by a collection of open sets has a finite subcover--that is, given a collection of open sets whose union contains $X,$ it is possible to choose a subcollection of finitely many open sets from the covering whose union still contains $X.$

In the absence of a metric, it is possible to recover many of the definitions and properties of metric spaces for arbitrary sets. The idea is, given a set $X,$ to specify a collection of open subsets (called a topology) satisfying the following axioms:

• The empty set and $X$ are open.

• An infinite union of open sets is open; a finite intersection of open sets is open.

These are, in a sense, the fundamental properties of open sets. These axioms allow for broad generalizations of open sets to contexts in which there is no natural metric. Indeed, there are some important examples of topologies in mathematics which do not come from metrics, including the Zariski topology in algebraic geometry.

1. Pred, . Open set - example. Retrieved January 24, 2009, from https://commons.wikimedia.org/wiki/File:Open_set_-_example.png

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Source: https://brilliant.org/wiki/open-sets/

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