Continuity and Open Sets

Openness and Continuity

In the previous tutorial, continuity was characterized through the preimages of closed sets. There is a parallel description using open sets: a continuous function pulls open sets back to sets that are open relative to its domain. The relative qualification matters whenever the domain is not all of $\mathbb{R}$. A set can be open within a domain even though it is not open in the entire real line.

This connection also gives a useful test for continuity. Instead of checking the epsilon-delta condition point by point, we can ask whether the preimage of every open set is relatively open. We will define relative openness, prove both directions of this characterization, and use examples to clarify what it does—and does not—say about images of sets.

Definition (Open and Relatively Open Sets): A set $U\subseteq\mathbb{R}$ is open if for every $y\in U$, there is an $r>0$ such that $(y-r,y+r)\subseteq U$. Given a domain $E\subseteq\mathbb{R}$, a set $A\subseteq E$ is open relative to $E$, or open in $E$, if for every $x\in A$, there is a $\delta>0$ such that $$ E\cap(x-\delta,x+\delta)\subseteq A. $$

The intersection with $E$ is essential: only points of the domain need to lie in $A$. Equivalently, $A$ is open relative to $E$ exactly when there is an open set $V\subseteq\mathbb{R}$ such that $A=E\cap V$. To see why, if $A$ is relatively open, take the union of the open intervals $(x-\delta_x,x+\delta_x)$ supplied by its definition for all $x\in A$. This union is open, and its intersection with $E$ is $A$. In the other direction, if $A=E\cap V$ and $V$ is open, every $x\in A$ has a small interval around it contained in $V$, so the interval’s intersection with $E$ lies in $A$.

For example, $[0,1)$ is open relative to $[0,3)$: it is $[0,3)\cap(-1,1)$. It is not open in $\mathbb{R}$, since no interval around $0$ is contained in $[0,1)$. Relative openness allows the domain to supply a boundary: points outside the domain do not affect the condition.

Continuous Functions Pull Back Open Sets

If $f:E\to\mathbb{R}$ is continuous and $U$ is open in $\mathbb{R}$, consider an input $x$ whose output $f(x)$ lies in $U$. Openness gives a neighborhood around $f(x)$ that stays inside $U$. Continuity then ensures that inputs sufficiently close to $x$, within the domain, have outputs in that neighborhood. Thus $x$ has a relative neighborhood contained in the preimage.

Theorem (Open Preimages Under Continuous Functions): Let $E\subseteq\mathbb{R}$, let $f:E\to\mathbb{R}$ be continuous, and let $U\subseteq\mathbb{R}$ be open. Then $$ f^{-1}(U)=\{x\in E:f(x)\in U\} $$ is open relative to $E$.

Proof. If $f^{-1}(U)$ is empty, it is relatively open by the definition, which places a condition only on points in the set. Otherwise, let $x\in f^{-1}(U)$, so $f(x)\in U$. Since $U$ is open, there is an $\varepsilon>0$ such that $$ (f(x)-\varepsilon,f(x)+\varepsilon)\subseteq U. $$ Continuity of $f$ at $x$ gives a $\delta>0$ such that, whenever $z\in E$ and $|z-x|<\delta$, we have $|f(z)-f(x)|<\varepsilon$. Consequently $f(z)\in U$. In other words, $$ z\in E\cap(x-\delta,x+\delta)\quad\Longrightarrow\quad z\in f^{-1}(U). $$ Every point $x$ of $f^{-1}(U)$ therefore has a relative neighborhood contained in that preimage. It is open relative to $E$. $\square$

Worked Example: A Polynomial Preimage

Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=x^2$, and let $U=(4,9)$. The polynomial is continuous, and $U$ is open. Solving the condition $f(x)\in U$ gives

x\in f^{-1}((4,9)) \quad\Longleftrightarrow\quad 4<x^2<9 \quad\Longleftrightarrow\quad 2<|x|<3 \quad\Longleftrightarrow\quad x\in(-3,-2)\cup(2,3).

Thus $f^{-1}((4,9))=(-3,-2)\cup(2,3)$, which is open in $\mathbb{R}$, as the theorem predicts. The two intervals arise because both positive and negative inputs have squares between $4$ and $9$.

Worked Example: A Relatively Open Preimage

Take the domain $E=[0,3)$, let $f:E\to\mathbb{R}$ be the identity function $f(x)=x$, and choose the open set $U=(-1,1)$. The identity function is continuous, and its preimage is

f^{-1}((-1,1)) =\{x\in[0,3):-1<x<1\} =[0,1).

This set is relatively open in $E$, since $[0,1)=[0,3)\cap(-1,1)$. For instance, at the domain point $0$, the relative neighborhood $[0,3)\cap(-1,1/2)=[0,1/2)$ lies in $[0,1)$. The preimage is not open in $\mathbb{R}$, but the theorem does not claim that it must be: its conclusion is openness relative to the domain.

The Open-Set Criterion for Continuity

The implication in the theorem has a converse. If every open set in the codomain has a relatively open preimage, then $f$ is continuous. The reason is that we can choose an open interval around $f(a)$ whose radius is any prescribed output tolerance. Relative openness of its preimage supplies the input neighborhood required by the epsilon-delta definition.

Theorem (Open-Set Criterion for Continuity): Let $E\subseteq\mathbb{R}$ and $f:E\to\mathbb{R}$. Then $f$ is continuous on $E$ if and only if $f^{-1}(U)$ is open relative to $E$ for every open set $U\subseteq\mathbb{R}$.

Proof. If $f$ is continuous on $E$, the Open Preimages Under Continuous Functions Theorem shows that $f^{-1}(U)$ is relatively open for every open $U\subseteq\mathbb{R}$.

Conversely, suppose the preimage of every open subset of $\mathbb{R}$ is open relative to $E$. Fix $a\in E$ and $\varepsilon>0$, and consider the open interval $$ U=(f(a)-\varepsilon,f(a)+\varepsilon). $$ Because $f(a)\in U$, we have $a\in f^{-1}(U)$. By the assumption, this preimage is open relative to $E$. Therefore there is a $\delta>0$ such that $$ E\cap(a-\delta,a+\delta)\subseteq f^{-1}(U). $$ Whenever $x\in E$ and $|x-a|<\delta$, it follows that $x\in f^{-1}(U)$, so $f(x)\in U$. Hence $|f(x)-f(a)|<\varepsilon$. This is precisely the epsilon-delta condition for continuity at $a$. Since $a$ was arbitrary, $f$ is continuous on $E$. $\square$

Worked Example: A Discontinuity Detected by an Open Set

Define $q:\mathbb{R}\to\mathbb{R}$ by $q(x)=0$ for $x\leq0$ and $q(x)=1$ for $x>0$. Take the open set $U=(-1/2,1/2)$. Since $q(x)=0$ exactly when $x\leq0$, and $0\in U$ while $1\notin U$, we get

q^{-1}((-1/2,1/2))=(-\infty,0].

This preimage is not open in $\mathbb{R}$: every interval around $0$ contains positive points, which are not in $(-\infty,0]$. The Open-Set Criterion therefore confirms that $q$ is not continuous on $\mathbb{R}$. In particular, $q(0)=0$, whereas $q(1/n)=1$ for every positive integer $n$, so the values along $1/n\to0$ do not converge to $q(0)$.

Preimages Are Not Images

A common pitfall is to reverse the direction of the open-preimage theorem. Continuity says that preimages of open sets are relatively open; it does not say that continuous functions send open sets to open sets. For example, $f(x)=x^2$ is continuous on $\mathbb{R}$, and $(-1,1)$ is open, but $$ f((-1,1))=[0,1) $$ is not open in $\mathbb{R}$. Indeed, the image includes $0$, and every interval around $0$ contains negative numbers outside the image. The preimage and image operations have different roles, so the theorem about preimages cannot be transferred to images.

The set-based criterion is useful when a direct epsilon-delta proof is inconvenient. To prove continuity, it may be easier to describe the solution set to $f(x)\in U$ for a general open $U$ and establish that this set is relatively open. Conversely, to disprove continuity, it is enough to find one open set whose preimage fails relative openness. In either direction, keeping track of the domain prevents errors at its boundary: the relevant neighborhoods are intersected with $E$, not tested against points that are not inputs of $f$.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What condition must hold around each point of a set $A$ for $A$ to be open relative to $E$?
Why does continuity make the preimage of an open set relatively open?
How does relative openness of all open-set preimages imply the epsilon-delta condition?
For the identity function on $[0,3)$, why is $[0,1)$ relatively open even though it is not open in $\mathbb{R}$?
What distinction between preimages and images is illustrated by $x\mapsto x^2$ on $(-1,1)$?

What You'll Learn