The Usual Metric Induces the Standard Topology on R

From Distance to Intervals

The usual distance between real numbers is $d(x,y)=|x-y|$. Measuring distance from a fixed point $x$ gives a simple description of an open ball: the points less than a prescribed distance from $x$ lie between two endpoints. This observation connects the metric description of openness with the interval description used for the standard topology on $\mathbb{R}$.

We will prove the connection in both directions. First, every metric-open set is a union of open intervals. Conversely, every union of open intervals is metric-open. The two descriptions therefore specify exactly the same open sets, not merely similar collections of sets.

Definition: For $x\in\mathbb{R}$ and $r>0$, the open ball with center $x$ and radius $r$, for the usual metric $d(x,y)=|x-y|$, is $B_d(x,r)=\{y\in\mathbb{R}:|y-x|<r\}$. A set $U\subseteq\mathbb{R}$ is metric-open if for every $x\in U$, there is some $r>0$ such that $B_d(x,r)\subseteq U$.

For comparison, let $\mathcal{T}_{\mathrm{std}}$ denote the standard topology on $\mathbb{R}$: its open sets are arbitrary unions of open intervals. An open interval may be bounded, such as $(a,b)$, or unbounded, such as $(a,\infty)$ or $\mathbb{R}$. The empty set is the union of no intervals and is open as well.

Open Balls Are Open Intervals

Proposition: For every $x\in\mathbb{R}$ and $r>0$, the usual-metric ball $B_d(x,r)$ is the interval $(x-r,x+r)$.

Proof. Let $y\in\mathbb{R}$. By the definition of the ball, $y\in B_d(x,r)$ exactly when $|y-x|<r$. For real numbers, $|y-x|<r$ is equivalent to $-r<y-x<r$. Adding $x$ to all three parts gives $x-r<y<x+r$, which is exactly the condition $y\in(x-r,x+r)$. Thus the two sets have the same elements. $\square$

The endpoints are not included because membership requires a strict distance inequality. The proposition also works in reverse: the bounded open interval $(a,b)$ is a ball with center $(a+b)/2$ and radius $(b-a)/2$. Indeed, for $a<b$, these choices give $x-r=a$ and $x+r=b$.

Worked Example: Finding a Ball as an Interval

Write $B_d(3,\frac{5}{2})$ as an interval and check whether $1$ and $6$ belong to it.

By the proposition, the left endpoint is $3-\frac{5}{2}=\frac{6}{2}-\frac{5}{2}=\frac{1}{2}$, and the right endpoint is $3+\frac{5}{2}=\frac{6}{2}+\frac{5}{2}=\frac{11}{2}$. Therefore $B_d(3,\frac{5}{2})=(\frac{1}{2},\frac{11}{2})$.

For $1$, the distance from the center is $|1-3|=2$, and $2<\frac{5}{2}$, so $1$ is in the ball. For $6$, the distance is $|6-3|=3$, and $3\not<\frac{5}{2}$, so $6$ is not in the ball.

Check: The interval test gives $\frac{1}{2}<1<\frac{11}{2}$, so $1$ is included, while $6>\frac{11}{2}$, so $6$ is excluded. This agrees with the distance test.

The Two Descriptions of Open Sets Agree

Theorem: A subset $U$ of $\mathbb{R}$ is open for the usual metric if and only if it is a union of open intervals. Equivalently, the usual metric induces the standard topology on $\mathbb{R}$.

Proof. Suppose first that $U$ is metric-open. For each $x\in U$, the definition of metric openness supplies a radius $r_x>0$ such that $B_d(x,r_x)\subseteq U$. By the proposition, $B_d(x,r_x)=(x-r_x,x+r_x)$, an open interval. Every point $x\in U$ belongs to its corresponding ball, so $U$ is the union of these intervals: $$ U=\bigcup_{x\in U}B_d(x,r_x). $$ If $U=\varnothing$, this is the union over no points, which is the empty set and is still a union of open intervals. Thus every metric-open set is a union of open intervals.

Conversely, suppose $U$ is a union of open intervals, and take any $x\in U$. Then $x$ belongs to at least one of the intervals in that union. We check the possible forms of that interval. If it is $(a,b)$, then $a<x<b$. The number $r=\min\{x-a,b-x\}$ is positive. If $y\in B_d(x,r)$, then $-r<y-x<r$, so $x-r<y<x+r$. Since $r\leq x-a$ and $r\leq b-x$, these inequalities imply $a\leq x-r<y<x+r\leq b$; hence $y\in(a,b)\subseteq U$.

If the interval is $(a,\infty)$, take $r=x-a>0$. For $y\in B_d(x,r)$, $y>x-r=a$, so $y\in(a,\infty)\subseteq U$. If it is $(-\infty,b)$, take $r=b-x>0$; then $y<x+r=b$, so $y\in(-\infty,b)\subseteq U$. If the interval is all of $\mathbb{R}$, any radius, for example $r=1$, gives a ball contained in it. In each case $x$ has a ball contained in $U$. Since this holds for every $x\in U$, $U$ is metric-open. The two collections of open sets are therefore equal, as claimed. $\square$

The proof uses two local facts: every point in a metric-open set has a ball around it lying in the set, and every point in an open interval has some positive distance to each finite endpoint of that interval. The second fact produces a suitable ball even when the interval itself is not centered at the point.

Worked Example: Decomposing a Bounded Interval into Balls

Show explicitly that $(1,4)$ is a union of usual-metric balls. For each $n\in\mathbb{N}$, consider the ball with center $\frac{5}{2}$ and radius $\frac{3}{2}-\frac{1}{n+2}$.

The radius is positive for every $n\geq1$: since $n+2\geq3$, we have $\frac{1}{n+2}\leq\frac{1}{3}$, and so $\frac{3}{2}-\frac{1}{n+2}\geq\frac{3}{2}-\frac{1}{3}=\frac{7}{6}>0$. The ball is the interval $$ B_d\left(\frac{5}{2},\frac{3}{2}-\frac{1}{n+2}\right) =\left(1+\frac{1}{n+2},\,4-\frac{1}{n+2}\right), $$ because $\frac{5}{2}-\frac{3}{2}=1$, $\frac{5}{2}+\frac{3}{2}=4$, and subtracting the smaller radius shifts the left endpoint right and the right endpoint left.

Every such interval lies in $(1,4)$, since its left endpoint is greater than $1$ and its right endpoint is less than $4$. To see that their union contains every $x\in(1,4)$, let $\delta=\min\{x-1,4-x\}$, which is positive. Choose $n\in\mathbb{N}$ large enough that $\frac{1}{n+2}<\delta$. Then $1+\frac{1}{n+2}<x<4-\frac{1}{n+2}$, so $x$ belongs to the $n$th interval.

Conclusion and check: The intervals are all contained in $(1,4)$, and every point of $(1,4)$ lies in one of them. Therefore $$ (1,4)=\bigcup_{n\in\mathbb{N}}B_d\left(\frac{5}{2},\frac{3}{2}-\frac{1}{n+2}\right). $$ For example, at $n=1$ the radius is $\frac{3}{2}-\frac{1}{3}=\frac{7}{6}$, giving $(\frac{4}{3},\frac{11}{3})$, which is indeed inside $(1,4)$.

Worked Example: A Disconnected, Unbounded Open Set

Verify directly that $U=(-2,1)\cup(4,\infty)$ is metric-open by writing it as a union of balls.

For the first interval, use balls centered at $-\frac{1}{2}$ with radii $\frac{3}{2}-\frac{1}{n+1}$, for $n\in\mathbb{N}$. These radii are positive because $n+1\geq2$, giving $\frac{3}{2}-\frac{1}{n+1}\geq\frac{3}{2}-\frac{1}{2}=1>0$. Their endpoints are $-\frac{1}{2}-(\frac{3}{2}-\frac{1}{n+1})=-2+\frac{1}{n+1}$ and $-\frac{1}{2}+(\frac{3}{2}-\frac{1}{n+1})=1-\frac{1}{n+1}$. Thus these balls give intervals inside $(-2,1)$, and as $n$ increases their endpoints approach $-2$ and $1$. More explicitly, for $x\in(-2,1)$, choose $n$ with $\frac{1}{n+1}<\min\{x+2,1-x\}$; then $x$ lies in the corresponding interval.

For the second interval, for each $n\in\mathbb{N}$ take the ball $B_d(4+n,n)$. Its interval form is $(4+n-n,4+n+n)=(4,4+2n)$. Each such interval is contained in $(4,\infty)$, and any $x>4$ belongs to one of them by choosing $n\in\mathbb{N}$ with $2n>x-4$. Consequently, $$ U=\left(\bigcup_{n\in\mathbb{N}}B_d\left(-\frac{1}{2},\frac{3}{2}-\frac{1}{n+1}\right)\right) \cup\left(\bigcup_{n\in\mathbb{N}}B_d(4+n,n)\right). $$

Check: Every ball in the first union is contained in $(-2,1)$, and every ball in the second is contained in $(4,\infty)$. The point $2$ is in neither interval and is not in any of the displayed balls. Thus the union is exactly $U$, including the gap $[1,4]$ between its two components.

Endpoint Strictness and a Common Mistake

The equality between balls and open intervals depends on the strict inequality in the ball definition. Confusing an open ball with a closed interval changes the set and can lead to an incorrect conclusion about openness.

Worked Example: Why a Closed Interval Is Not Open

Consider $C=[-1,2]$. Every point $x$ with $-1<x<2$ has a positive distance to both endpoints. In particular, $r=\min\{x+1,2-x\}>0$ gives $B_d(x,r)\subseteq C$. But the endpoint $-1$ has no ball contained in $C$: for any $r>0$, the point $y=-1-\frac{r}{2}$ satisfies $|y-(-1)|=\frac{r}{2}<r$, so $y\in B_d(-1,r)$, while $y<-1$, so $y\notin C$. Thus $C$ is not metric-open.

Common mistake: One might write $B_d(-1,1)=[-2,0]$, then treat the ball as a closed interval. The definition instead gives $B_d(-1,1)=\{y:|y+1|<1\}=(-2,0)$. For example, $|-2+1|=1$, which is not less than $1$, so $-2$ is not in the ball; likewise $0$ is excluded. The strict endpoints matter: a ball is an open interval, and a closed interval with an endpoint cannot be open in the usual topology.

This identification is useful whenever a problem is phrased in one language but easier to solve in the other. A metric condition can be replaced by interval inequalities, while interval-based reasoning about open sets can be converted into balls around each point. The equivalence also confirms that referring to the standard topology on $\mathbb{R}$ or to the topology induced by the usual metric specifies the same open sets.

Check Your Understanding

For each question, use the ball definition or the interval characterization as appropriate.

Write $B_d(-2,3)$ as an open interval. Are $-5$ and $0$ in the ball?
For $a<b$, find a center and positive radius whose ball is exactly $(a,b)$.
Give a positive radius $r$ such that $B_d(x,r)\subseteq(a,b)$, assuming $a<x<b$.
Explain why $(-\infty,3)$ is metric-open by giving a ball around each of its points.
Is $[0,1)$ metric-open? Identify a point that determines the answer and justify it using the ball definition.