Open and Closed Sets in Standard, Discrete, and Induced Metrics

Hello! Last time, you checked whether a proposed distance really is a metric. Now we will use that distance to decide which subsets are open and closed. The central question is local: from each point of a set, can you move a sufficiently small distance without leaving it?

We will answer that question in the real line and plane, in a discrete space, and in subsets equipped with an induced metric. The last case is especially important: whether a set is open or closed depends not only on the set, but also on the space in which you are considering it.

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Balls turn distance into openness

In a metric space $(X,d)$, the open ball of radius $r>0$ centered at $x\in X$ is

$$B_X(x,r)=\{y\in X:d(x,y)<r\}.$$

The subscript is a reminder that only points of $X$ can belong to this ball. A subset $U\subseteq X$ is open in $X$ if every $x\in U$ has some radius $r_x>0$ for which $B_X(x,r_x)\subseteq U$. The radius may depend on the point; it need not work uniformly across $U$.

Watch the opening of Functional Analysis 3 | Open and Closed Sets from The Bright Side of Mathematics for the ball definition and its visual interpretation. Then watch the short segment giving the definition of closedness; we will develop that definition from balls below.

The picture suggests the definition, but the metric makes it precise. On $\mathbb{R}$ with $d(x,y)=|x-y|$, a ball is the interval

$$B_{\mathbb{R}}(x,r)=(x-r,x+r).$$

If $a<x<b$, choosing $r=\min \{x-a,b-x\}>0$ keeps this entire interval inside $(a,b)$. Thus $(a,b)$ is open. In contrast, $[a,b)$ is not open in $\mathbb{R}$: it contains $a$, and every ball around $a$ contains points less than $a$.

The shape of a ball changes with the metric. In ${\mathbb{R}}^2$, a Euclidean ball is a disk; for the maximum metric from the previous lesson it is the inside of an axis-aligned square; for the taxicab metric it is the inside of a diamond. In every case, the test for openness is the same: can you fit an appropriate ball around each point of the proposed set?

Read the ball examples in Notes on Metric Spaces, including its first example of balls in a subspace. This revisits metrics you already know without repeating their axiom checks.

One fact underlies all these examples: every open ball is itself an open set, in any metric space. This is a useful short proof to read closely.

Here is the proof’s key move in our notation. Given $y\in B_X(x,r)$, there is a positive margin $s=r-d(x,y)$. If $z\in B_X(y,s)$, then

$$d(x,z)\le d(x,y)+d(y,z)<d(x,y)+s=r.$$

Consequently $B_X(y,s)\subseteq B_X(x,r)$. Notice the order of the reasoning: choose an arbitrary point $y$ in the original ball, construct a radius that works for that point, and check the containment for every $z$ in the smaller ball.

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Closed means an open complement

A set $F\subseteq X$ is closed in $X$ when $X\setminus F$ is open in $X$. Thus you test closedness by taking an arbitrary point outside $F$ and finding a ball around it that misses $F$.

For instance, $[a,b]$ is closed in $\mathbb{R}$. If $x<a$, the positive radius $a-x$ gives a ball that misses $[a,b]$; if $x>b$, use $x-b$. These two cases cover its complement. By comparison, $(a,b]$ is neither open nor closed in $\mathbb{R}$: no ball centered at its included endpoint $b$ stays inside it, while its complement contains $a$ but no ball around $a$ stays outside it.

The same argument works for a closed ball

$${\overline{B}}_X(x,r)=\{z\in X:d(x,z)\le r\}.$$

To check it is closed, take $y$ outside it, so $d(x,y)>r$. Put $s=d(x,y)-r>0$. For any $z\in B_X(y,s)$, the triangle inequality gives

$$d(x,z)\ge d(x,y)-d(y,z)>d(x,y)-s=r.$$

Hence the ball around $y$ misses ${\overline{B}}_X(x,r)$, proving its complement open. This is the same margin idea as before, applied on the outside.

Do not interpret “closed” as “not open.” The empty set and $X$ are both open and closed in $X$: openness of the empty set has no points to check, every ball in $X$ lies in $X$, and each is the other’s complement. A set can also be neither, as $(a,b]$ shows.

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The discrete metric changes the answer

Recall that under the discrete metric, distinct points have distance $1$. For every $x\in X$,

$$B_X(x,1)=\{x\}.$$

Take any subset $A\subseteq X$. Every $x\in A$ has the ball $B_X(x,1)=\{x\}\subseteq A$, so $A$ is open. The same argument applies to $X\setminus A$, making that complement open too. Every subset of a discrete metric space is therefore both open and closed.

This applies even if the underlying set is $\mathbb{R}$. The singleton $\{0\}$ is not open in $\mathbb{R}$ with the usual metric, because every usual interval around $0$ contains other real numbers. It is open on the same set of real numbers with the discrete metric. Knowing the points alone does not determine the open sets; you need the metric.

The strict inequality in the phrase open ball also does not mean that such a ball cannot be closed. In a discrete space, $B_X(x,1)=\{x\}$ is both.

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Induced metrics: which space contains the ball?

Suppose $Y\subseteq X$ and $Y$ carries the induced metric $d_Y$, obtained by restricting $d$ to pairs of points in $Y$. Directly from the ball definition, for $y\in Y$,

$$B_Y(y,r)=B_X(y,r)\cap Y.$$

This identity is the safest starting point for a subspace question. It says that a ball in $Y$ can be shorter, one-sided, or otherwise truncated compared with the ball in $X$. For example, when $Y=\mathbb{Q}\subseteq \mathbb{R}$, the induced-metric ball is $(q-r,q+r)\cap \mathbb{Q}$, not the entire real interval.

There is a useful general rule:

A set $U\subseteq Y$ is open in $Y$ exactly when $U=Y\cap V$ for some open $V\subseteq X$. Likewise, $F\subseteq Y$ is closed in $Y$ exactly when $F=Y\cap C$ for some closed $C\subseteq X$.

Study the subspace theorem and its proof in LibreTexts’ 3.8: Open and Closed Sets. Neighborhoods. The proof is worth reading as a model of how a statement about all subspace-open sets is built from the ball identity above.

For a quick concrete instance, let $Y=[0,1]$. The set $[0,\frac{1}{2})$ is open in $Y$, since it equals $Y\cap (-1,\frac{1}{2})$, with $(-1,\frac{1}{2})$ open in $\mathbb{R}$. It is not open in $\mathbb{R}$: the point $0$ has no real ball contained in it. Meanwhile, $Y$ itself is both open and closed in $Y$, though it is not open in $\mathbb{R}$.

An ambient interval can even become a nontrivial set that is both open and closed in a subspace. Return to the same video for an example with a gap in the surrounding space.

In that example, take $X=(1,3]\cup (4,\infty )$ with the usual distance restricted from $\mathbb{R}$, and $A=(1,3]$. At $3$, the ball $B_X(3,1)$ contains no point of the other component: its points are restricted to $X$, and points greater than $4$ are more than $1$ away. The other points of $A$ also have sufficiently small balls inside $A$, so $A$ is open in $X$. Its complement $(4,\infty )$ is open in $X$, so $A$ is closed there as well. Yet $(1,3]$, considered as a subset of $\mathbb{R}$, is neither open nor closed.

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Takeaways

To determine whether $A\subseteq X$ is open, start with an arbitrary point in $A$ and find a ball contained in $A$. To determine whether it is closed, start with an arbitrary point outside $A$ and find a ball that misses $A$. A failed test needs only one point for which every radius fails; a successful test must account for every relevant point.

Always name the metric space first. Discrete balls make every subset open and closed, while induced-metric balls are intersections with the surrounding space’s balls. Next, we will use these same local ball tests to compute the interior, closure, and boundary of concrete sets.