Continuity and Closure Preservation

Continuity and What It Does to Closures

Continuity controls how a map behaves near every point, but it can also be expressed using entire subsets. The key comparison is between the image of a set’s closure and the closure of its image. Continuity guarantees a particular containment for every subset of the domain. Conversely, if that containment holds for every subset, it forces continuity.

Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be topological spaces, and let $f:X\to Y$. Recall that $f$ is continuous if $f^{-1}(V)$ is open in $X$ for every open set $V$ in $Y$. A set is closed if its complement is open. The closure $\overline{A}$ of $A\subseteq X$ is the intersection of all closed subsets of $X$ containing $A$; it is the smallest closed set containing $A$.

Theorem (Continuity and closure preservation): A map $f:X\to Y$ is continuous if and only if, for every subset $A\subseteq X$, $$ f(\overline{A})\subseteq\overline{f(A)}. $$ The closure on the left is taken in $X$, and the closure on the right is taken in $Y$.

The condition is an inclusion, not necessarily an equality. Also, it must hold for every subset $A$ to characterize continuity; checking a single convenient set is not enough.

A Neighborhood Test for Closure

We will use the following test. An open neighborhood of $x$ means an open set containing $x$.

Lemma: For $A\subseteq X$, a point $x$ lies in $\overline{A}$ if and only if every open neighborhood of $x$ intersects $A$.

Proof. Suppose $x\in\overline{A}$, and let $U$ be an open neighborhood of $x$. If $U\cap A=\varnothing$, then $X\setminus U$ is closed and contains $A$. The definition of closure would give $\overline{A}\subseteq X\setminus U$, contradicting $x\in U\cap\overline{A}$. Thus $U\cap A\ne\varnothing$. Conversely, suppose every open neighborhood of $x$ intersects $A$. If $x\notin\overline{A}$, then $X\setminus\overline{A}$ is an open neighborhood of $x$, since $\overline{A}$ is closed. It misses $A$, because $A\subseteq\overline{A}$, a contradiction. Therefore $x\in\overline{A}$. $\square$

Proof of the Criterion

Proof. First suppose $f$ is continuous. Let $A\subseteq X$, and choose any $y\in f(\overline{A})$. There is $x\in\overline{A}$ with $f(x)=y$. To show $y\in\overline{f(A)}$, take any open neighborhood $V$ of $y$ in $Y$. Continuity makes $f^{-1}(V)$ open in $X$, and it contains $x$. By the closure test, $f^{-1}(V)$ intersects $A$. Choose $a\in f^{-1}(V)\cap A$. Then $f(a)\in V\cap f(A)$. Every open neighborhood $V$ of $y$ therefore intersects $f(A)$, so $y\in\overline{f(A)}$. This proves $f(\overline{A})\subseteq\overline{f(A)}$.

Conversely, suppose $f(\overline{A})\subseteq\overline{f(A)}$ for every $A\subseteq X$. Let $C$ be any closed subset of $Y$, and set $A=f^{-1}(C)$. Since $f(A)\subseteq C$ and $C$ is closed, the definition of closure gives $\overline{f(A)}\subseteq C$. The assumed condition now gives $f(\overline{A})\subseteq C$, so $\overline{A}\subseteq f^{-1}(C)=A$. On the other hand, $A\subseteq\overline{A}$, hence $A=\overline{A}$. The closure is closed, so $f^{-1}(C)=A$ is closed.

Thus preimages of closed sets are closed. For any open $V\subseteq Y$, its complement $Y\setminus V$ is closed, and $f^{-1}(Y\setminus V)=X\setminus f^{-1}(V)$. Since the left side is closed, $f^{-1}(V)$ is open. This is continuity. The argument includes $A=\varnothing$, for which both closures are empty, and it also applies when $X$ is empty. $\square$

Using the Inclusion

Worked Example: An Affine Map of the Real Line

Let $X=Y=\mathbb{R}$ with the standard topology, let $f(x)=2x+1$, and take $A=(0,1]\cup\{3\}$. First, $f$ is continuous: the preimage of an open interval $(u,v)$ is the open interval $((u-1)/2,(v-1)/2)$, and open intervals form a basis for the standard topology.

The closure of $A$ is $[0,1]\cup\{3\}$. The points of $(0,1]$ and the point $3$ already belong to $A$; $0$ is approached by points in $(0,1]$; and any real point outside $[0,1]\cup\{3\}$ has an open interval around it missing $A$. Hence $f(\overline{A})=[1,3]\cup\{7\}$.

For the image itself, $f((0,1])=(1,3]$ and $f(3)=7$, so $f(A)=(1,3]\cup\{7\}$. Its closure is $[1,3]\cup\{7\}$: the missing endpoint $1$ is approached from within $(1,3]$, while $7$ is an isolated point of this set. Therefore $\overline{f(A)}=[1,3]\cup\{7\}=f(\overline{A})$.

Check: the endpoint $1$ on the left comes from $f(0)=1$, even though $0\notin A$. The endpoint $3$ comes from $f(1)=3$, and $7$ comes from $f(3)=7$. The two computed sets agree, as the criterion permits; continuity requires the inclusion for every subset, not just this one.

Worked Example: A Continuous Inclusion with Strict Containment

Let $X=(0,1)$ with its subspace topology from the standard real line, let $Y=\mathbb{R}$ with the standard topology, and define $f:X\to Y$ by $f(x)=x$. This inclusion is continuous: for every open $V\subseteq Y$, $f^{-1}(V)=V\cap(0,1)$, which is open in the subspace topology.

Take $A=(0,1/2)\subseteq X$. Its closure in $X$ is $(0,1/2]$. Indeed, every $x$ with $0<x\leq 1/2$ has every relative open neighborhood intersecting $A$, while for $x>1/2$ in $X$, a sufficiently small relative interval around $x$ misses $A$. The endpoint $0$ is not a point of $X$, so it cannot belong to this closure. Thus $f(\overline{A})=(0,1/2]$.

In the codomain, $f(A)=(0,1/2)$ has closure $[0,1/2]$. The point $0$ lies in this closure because every standard open interval around $0$ intersects $(0,1/2)$; points outside $[0,1/2]$ have open intervals missing it. Consequently, $$ f(\overline{A})=(0,1/2]\subsetneq[0,1/2]=\overline{f(A)}. $$ The containment is strict because $0$ belongs to the right-hand set but not the left-hand set.

Check: every point of $(0,1/2]$ belongs to $[0,1/2]$, so the required inclusion holds. The difference between the two sets comes from taking closure in different spaces: $0$ is a limit point in $Y$, but it is not even a point of $X$.

Diagnosing Discontinuous Maps

Worked Example: Identity Between Two Finite Topologies

Let $X=\{a,b\}$ have topology $\mathcal{T}_X=\{\varnothing,\{a\},X\}$, and let $Y=\{a,b\}$ have topology $\mathcal{T}_Y=\{\varnothing,\{b\},Y\}$. Define $f:X\to Y$ to be the identity map, and take $A=\{a\}$.

The closed sets in $X$ are $X,\{b\},\varnothing$. The only one containing $A$ is $X$, so $\overline{A}=X$ in $X$, and $f(\overline{A})=X$. The closed sets in $Y$ are $Y,\{a\},\varnothing$, so $\overline{f(A)}=\overline{\{a\}}=\{a\}$ in $Y$. Therefore $f(\overline{A})=\{a,b\}\not\subseteq\{a\}=\overline{f(A)}$.

This failure proves that $f$ is not continuous. Directly, $\{b\}$ is open in $Y$, but $f^{-1}(\{b\})=\{b\}$ is not open in $X$. Both checks give the same conclusion.

Worked Example: The Standard Topology Mapped to the Discrete Topology

Let $X=\mathbb{R}$ with the standard topology and $Y=\mathbb{R}$ with the discrete topology, in which every subset is open. Let $f:X\to Y$ be the identity and take $A=(0,1)$. In $X$, $\overline{A}=[0,1]$. In the discrete space $Y$, every subset is closed as well as open, so $\overline{f(A)}=f(A)=(0,1)$. Hence $f(\overline{A})=[0,1]\not\subseteq(0,1)=\overline{f(A)}$.

The failure occurs at both endpoints: $0$ and $1$ lie in $f(\overline{A})$ but not in $\overline{f(A)}$. The map is indeed not continuous, since the open singleton $\{0\}$ in $Y$ has preimage $\{0\}$, which is not open in the standard topology on $X$.

Common pitfall. For this same map, testing only $A=\{0\}$ would give $f(\overline{A})=\{0\}=\overline{f(A)}$, because $\{0\}$ is closed in the standard real line and every set is closed in the discrete topology. That calculation does not establish continuity. The theorem requires the condition for every subset, and $A=(0,1)$ supplies a counterexample.

How to Read the Criterion

The theorem provides a practical way to disprove continuity: find a subset $A\subseteq X$ and a point $x\in\overline{A}$ for which $f(x)\notin\overline{f(A)}$. The two finite steps are to compute the closure in the domain, apply $f$, and compare that image with the closure of $f(A)$ in the codomain. A failed inclusion is conclusive; one successful computation is not.

For a continuous map, points that lie in the closure of $A$ must map to points that lie in the closure of $f(A)$. The converse proof explains why this statement captures the whole definition: applying it to the preimage of an arbitrary closed set forces that preimage to be closed. The choice of topology on each space matters at every closure calculation, as the examples show.

Check Your Understanding

For each question, state which topology is used for each closure you compute.

In the standard topology on $\mathbb{R}$, let $f(x)=x+2$ and $A=[0,2)$. Compute $f(\overline{A})$ and $\overline{f(A)}$.
For the identity map from the standard real line to the discrete real line, why does $A=\{0\}$ not detect discontinuity, while $A=(0,1)$ does?
In the finite-space example, compute $\overline{\{b\}}$ in $X$ and in $Y$. Does that subset violate the closure-preservation condition?
Explain why the converse proof uses the particular subset $A=f^{-1}(C)$ when $C$ is closed in $Y$.
Can equality $f(\overline{A})=\overline{f(A)}$ for one subset $A$ prove that $f$ is continuous? Give the reason for your answer.