Relative Neighborhoods and Local Tests

Neighborhoods Inside a Subspace

When a set $A$ is viewed as a space in its own right, its open sets need not be open in the surrounding space $X$. Instead, they are obtained by restricting ambient open sets to $A$. This gives a local way to test openness: around each point of a relatively open set, there must be an ambient open neighborhood whose part inside $A$ stays in the set.

Let $(X,\mathcal{T})$ be a topological space and $A\subseteq X$. The subspace topology on $A$ consists of the sets $A\cap U$ for $U\in\mathcal{T}$. Thus a set $G\subseteq A$ is open in $A$ precisely when it is the intersection of $A$ with some open set of $X$. An open neighborhood in $A$ of $x\in A$ means a set open in this subspace topology that contains $x$.

Definition: A set $N\subseteq A$ is a neighborhood of $x$ relative to $A$ if it contains an open neighborhood of $x$ in $A$. If $N$ is itself open in $A$ and contains $x$, it is an open neighborhood relative to $A$. For an ambient open set $V\subseteq X$ containing $x$, the set $A\cap V$ is an open neighborhood of $x$ relative to $A$.

A relative neighborhood need not itself be open, just as an ambient neighborhood need not itself be open. The useful point is that every relative neighborhood contains one of the form $A\cap V$, where $V$ is ambient open and contains the point.

Proposition: For $x\in A$ and $N\subseteq A$, $N$ is a neighborhood of $x$ relative to $A$ if and only if there is an open set $V\in\mathcal{T}$ such that $x\in V$ and $A\cap V\subseteq N$. If $N$ is open in $A$, this condition can be strengthened to $N=A\cap V$.

Proof. Suppose first that $N$ is a relative neighborhood of $x$. By definition, it contains an open neighborhood $G$ of $x$ in $A$. The subspace topology gives an ambient open set $V$ such that $G=A\cap V$. Since $x\in G$, we have $x\in V$, and $A\cap V=G\subseteq N$. Conversely, if such an ambient open $V$ exists, then $A\cap V$ is open in $A$, contains $x$, and is contained in $N$, so $N$ is a relative neighborhood. If $N$ is open in $A$, the definition of the subspace topology gives $N=A\cap V$ for some ambient open $V$; because $x\in N$, this $V$ contains $x$. $\square$

The Local Test for Relative Openness

Theorem (Local test): A set $G\subseteq A$ is open in the subspace $A$ if and only if for every $x\in G$ there is an ambient open set $V_x\subseteq X$ such that $x\in V_x$ and $A\cap V_x\subseteq G$.

Proof. Suppose $G$ is open in $A$. By the definition of the subspace topology, there is an ambient open set $U$ with $G=A\cap U$. For any $x\in G$, we have $x\in U$, and $A\cap U=G$, so $V_x=U$ satisfies the condition. Conversely, suppose the stated condition holds. For each $x\in G$, choose an ambient open $V_x$ with $x\in V_x$ and $A\cap V_x\subseteq G$. Then $$ G=\bigcup_{x\in G}(A\cap V_x) =A\cap\left(\bigcup_{x\in G}V_x\right). $$ The first equality holds because each $x\in G$ belongs to $A\cap V_x$, while every $A\cap V_x$ is contained in $G$. The union $\bigcup_{x\in G}V_x$ is open in $X$, so the displayed representation shows that $G$ is open in $A$. If $G=\varnothing$, it is open in $A$ directly, and the same representation uses the empty union. $\square$

The local test says that the ambient neighborhood may include points outside $A$; only its intersection with $A$ must stay inside $G$. This is the essential difference from testing whether $G$ is open in $X$. For relative openness, points outside the subspace impose no restriction.

Worked Example: A Finite Subspace

Let $X=\{a,b,c\}$ have topology $\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}$, and take $A=\{b,c\}$. Determine the open sets in $A$, then test whether $\{b\}$ and $\{c\}$ are open in $A$.

Intersect each ambient open set with $A$: $\varnothing\cap A=\varnothing$, $\{a\}\cap A=\varnothing$, $\{a,b\}\cap A=\{b\}$, and $X\cap A=\{b,c\}$. Therefore the subspace topology is $\{\varnothing,\{b\},A\}$, so $\{b\}$ is open in $A$ and $\{c\}$ is not.

The local test verifies both conclusions. For $b$, choose the ambient open set $V=\{a,b\}$. It contains $b$, and $A\cap V=\{b\}\subseteq\{b\}$. For $c$, the only ambient open set containing $c$ is $X$, and $A\cap X=A$, which is not contained in $\{c\}$. Thus no suitable ambient neighborhood exists at $c$.

Check: the only relative open set containing $c$ is $A$, so $\{c\}$ cannot be open in $A$. This agrees with the local test.

The Open-Ball Test in Metric Spaces

Now let $(X,d)$ be a metric space, and give $A\subseteq X$ the subspace topology. Write $B_d(x,r)=\{y\in X:d(x,y)<r\}$ for the ambient open ball of radius $r>0$. Since the open balls form a basis for the metric topology, the local test has a metric form.

Theorem (Open-ball test): A set $G\subseteq A$ is open in $A$ if and only if for every $x\in G$ there is $r>0$ such that $B_d(x,r)\cap A\subseteq G$.

Proof. Suppose $G$ is open in $A$. By the local test, for each $x\in G$ there is an ambient open set $V$ containing $x$ with $A\cap V\subseteq G$. Because $V$ is open in a metric space, there is $r>0$ such that $B_d(x,r)\subseteq V$. Hence $B_d(x,r)\cap A\subseteq V\cap A\subseteq G$. Conversely, if such a radius exists at every $x\in G$, then each ambient ball $B_d(x,r)$ is open, contains $x$, and has intersection with $A$ contained in $G$. The local test therefore shows that $G$ is open in $A$. When $G=\varnothing$, the condition has no points to check and the empty set is open. $\square$

Worked Example: An Open Part of a Parabola

In $\mathbb{R}^2$ with the Euclidean metric, let $A=\{(t,t^2):t\in\mathbb{R}\}$ and let $G=\{(t,t^2):t<1\}$. Show that $G$ is open in $A$ using the ball test.

Fix $x=(t,t^2)\in G$, so $t<1$, and choose $r=(1-t)/2$, which is positive. If $y=(u,u^2)\in B(x,r)\cap A$, then $|u-t|\leq\|y-x\|<r$. In particular $u<t+r=t+(1-t)/2=(t+1)/2<1$. Therefore $y\in G$, so $B(x,r)\cap A\subseteq G$. This works for every $x\in G$, proving relative openness.

Check: the chosen radius is positive because $1-t>0$. The strict ball inequality gives $u<t+r$, and $t+r=(t+1)/2<1$, as required.

Worked Example: A Relative Set That Is Not Open

For the same $A\subseteq\mathbb{R}^2$, let $W=\{(t,t^2):t\leq1\}$. Use the ball test to show that $W$ is not open in $A$.

The point $p=(1,1)$ belongs to $W$. We show that every ambient ball centered at $p$ contains a point of $A\setminus W$. Let $r>0$, and set $\delta=\min(r/4,1/10)>0$. The point $q=(1+\delta,(1+\delta)^2)$ belongs to $A\setminus W$, since its first coordinate is strictly greater than $1$. Its distance from $p$ satisfies $$ \|q-p\|^2 =\delta^2+(2\delta+\delta^2)^2 \leq \delta^2+(2.1\delta)^2 =5.41\delta^2 <9\delta^2. $$ Here $2\delta+\delta^2\leq2.1\delta$ follows from $\delta\leq1/10$, and $5.41<9$. Thus $\|q-p\|<3\delta$. If $r\leq 2/5$, then $\delta=r/4$, so $3\delta=3r/4<r$. If $r>2/5$, then $\delta=1/10$, so $3\delta=3/10<r$. In either case $q\in B(p,r)\cap A$ but $q\notin W$.

No radius at $p$ can satisfy $B(p,r)\cap A\subseteq W$. The open-ball test therefore shows that $W$ is not open in $A$.

Worked Example: An Isolated Point in a Subspace

Let $X=\mathbb{R}^2$ with the Euclidean metric, let $A=\{(n,0):n\in\mathbb{Z}\}$, and let $G=\{(0,0)\}$. Show that $G$ is open in $A$.

Choose $r=1/2$. For each integer $n\ne0$, $|n|\geq1$, so the Euclidean distance from $(n,0)$ to $(0,0)$ is $|n|\geq1>1/2$. The point $(0,0)$ has distance $0<1/2$ from itself. Consequently $B((0,0),1/2)\cap A=\{(0,0)\}=G$, which proves that $G$ is open in $A$.

Check: the ball uses the strict condition $d(y,(0,0))<1/2$. Every other point of $A$ has distance at least $1$, so none is included.

A Common Pitfall

Common mistake: “A subset $G\subseteq A$ is open in $A$ only if it is open in $X$.” This incorrectly replaces relative openness with ambient openness. For example, in $\mathbb{R}$ with its standard topology, take $A=\{0\}$ and $G=\{0\}$. The set $G$ is open in $A$, because $G=A\cap(-1,1)$. But $G$ is not open in $\mathbb{R}$: every open interval around $0$ contains nonzero real numbers and is therefore not contained in $\{0\}$.

The corrected test is local and relative: for each $x\in G$, find an ambient open neighborhood of $x$ whose intersection with $A$ lies inside $G$. In a metric space, it is enough to find an ambient open ball with that property. The ball itself need not lie in $G$, because it may contain points outside $A$.

Question	Local test
Is $G\subseteq A$ open in $A$?	For each $x\in G$, find ambient open $V\ni x$ with $A\cap V\subseteq G$.
Is $N\subseteq A$ a relative neighborhood of $x$?	Find ambient open $V\ni x$ with $A\cap V\subseteq N$.
Is $G\subseteq A$ open when $X$ is metric?	For each $x\in G$, find $r>0$ with $B_d(x,r)\cap A\subseteq G$.

Check Your Understanding

For each question, identify the ambient space and the subspace before applying the local test.

Let $A=(0,2)\subseteq\mathbb{R}$. Is $(0,1)$ open in $A$? Give an ambient open set that exhibits it.
In $\mathbb{R}$, let $A=\{0\}\cup\{2,3,4,\ldots\}$. Is $\{0\}$ open in $A$? Find a radius or explain why none works.
State the difference between an open neighborhood relative to $A$ and a general neighborhood relative to $A$.
In a metric space, explain why an ambient ball must be intersected with $A$ in the relative openness test.
Give an example of a set open in a subspace but not open in its ambient space, and verify both claims.

Question	Local test
Is \(G\subseteq A\) open in \(A\)?	For each \(x\in G\), find ambient open \(V\ni x\) with \(A\cap V\subseteq G\).
Is \(N\subseteq A\) a relative neighborhood of \(x\)?	Find ambient open \(V\ni x\) with \(A\cap V\subseteq N\).
Is \(G\subseteq A\) open when \(X\) is metric?	For each \(x\in G\), find \(r>0\) with \(B_d(x,r)\cap A\subseteq G\).