TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), with open and closed sets as in The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement and balls as in Open ball, closed ball and sphere in a metric space. Then:

Balls are open. $B (x, r)$ is open, for every $x \in X$ and every $r > 0$ .
Arbitrary unions. If $U$ is any collection of open subsets of $X$ , then $⋃ U$ is open.
Finite intersections. If $n \geq 1$ and $U_{0}, \dots, U_{n - 1}$ are open, then $U_{0} \cap \dots \cap U_{n - 1}$ is open.
Closed balls are closed. $\overset{ˉ}{B} (x, r)$ is closed, for every $x \in X$ and every $r > 0$ .

Together with the fact that $\emptyset$ and $X$ are open, recorded already in The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, claims 2 and 3 say that $T_{d}$ has exactly the closure properties that the word topology names.

Facts & Assumptions

Given: A metric space $(X, d)$ ; a point $x \in X$ and a real $r > 0$ ; a collection $U$ of open subsets of $X$ ; a natural $n \geq 1$ and open sets $U_{0}, \dots, U_{n - 1} \subseteq X$ .

[A1]

Open: $U$ is open when every $u \in U$ admits $t > 0$ with $B (u, t) \subseteq U$ ; closed means the complement is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[A2]

Balls: $B (u, t) = {w : d (u, w) < t}$ and $\overset{ˉ}{B} (u, t) = {w : d (u, w) \leq t}$ , and $B (u, t) \subseteq B (u, t^{'})$ whenever $0 < t \leq t^{'}$ (Open ball, closed ball and sphere in a metric space).

[L1]

Triangle inequality and symmetry of $d$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

Reverse triangle inequality: $∣ d (a, c) - d (b, c) ∣ \leq d (a, b)$ , so in particular $d (a, c) - d (b, c) \leq d (a, b)$ (The reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space).

[L3]

A nonempty finite set of reals has a minimum, which belongs to the set and is a lower bound of it (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L4]

Order arithmetic: a constant may be added to both sides of an inequality and inequalities may be chained by transitivity, in the strict form of Order is preserved by adding a constant and by adding inequalities and, with the case of equality settled by totality, in the nonstrict form (Ordered field, Complete ordered field (least-upper-bound property)); and by trichotomy $a < b$ and $b \leq a$ cannot both hold.

Proof

technique · direct

1.1A1A2L1L4

Claim 1: let $y \in B (x, r)$ , so $d (x, y) < r$ , and put $s : = r - d (x, y) > 0$ ; for $z \in B (y, s)$ the triangle inequality gives $d (x, z) \leq d (x, y) + d (y, z) < d (x, y) + s = r$ , so $B (y, s) \subseteq B (x, r)$ , and since $y$ was arbitrary $B (x, r)$ is open.

1.2A1

Claim 2: let $y \in ⋃ U$ , so $y \in U$ for some $U \in U$ ; as $U$ is open there is $t > 0$ with $B (y, t) \subseteq U \subseteq ⋃ U$ , and since $y$ was arbitrary the union is open.

1.3A1choose

Claim 3: let $y \in U_{0} \cap \dots \cap U_{n - 1}$ and for each $k < n$ pick $t_{k} > 0$ with $B (y, t_{k}) \subseteq U_{k}$ , which is possible because each $U_{k}$ is open and $y$ lies in it.

1.4A2L1L2L4

Claim 4: let $y \in X ∖ \overset{ˉ}{B} (x, r)$ , so $d (x, y) > r$ , and put $s : = d (x, y) - r > 0$ ; for $z \in B (y, s)$ the reverse triangle inequality applied to the points $y, z, x$ gives $d (y, x) - d (z, x) \leq d (y, z) < s$ , hence $d (z, x) > d (y, x) - s = r$ , so $d (x, z) > r$ by symmetry and $z \notin \overset{ˉ}{B} (x, r)$ .

2.1step 1.3L3

Since $n \geq 1$ , the set ${t_{0}, \dots, t_{n - 1}}$ is a nonempty finite set of reals, so $t : = \min {t_{0}, \dots, t_{n - 1}}$ exists, equals some $t_{j}$ and is therefore $> 0$ , and satisfies $t \leq t_{k}$ for every $k < n$ .

2.2step 1.4A1

Step 1.4 shows $B (y, s) \subseteq X ∖ \overset{ˉ}{B} (x, r)$ for the $y$ and $s$ chosen there, and $y$ was an arbitrary point of $X ∖ \overset{ˉ}{B} (x, r)$ ; hence $X ∖ \overset{ˉ}{B} (x, r)$ is open and $\overset{ˉ}{B} (x, r)$ is closed, which is claim 4.

3.1step 2.1A1A2

By step 2.1, $B (y, t) \subseteq B (y, t_{k}) \subseteq U_{k}$ for every $k < n$ , so $B (y, t) \subseteq U_{0} \cap \dots \cap U_{n - 1}$ ; as $y$ was arbitrary that intersection is open, which is claim 3.

4.1step 1.1step 1.2step 2.2step 3.1∎

Claims 1, 2, 3 and 4 are established by steps 1.1, 1.2, 3.1 and 2.2 respectively.

Remarks

Finiteness in claim 3 is essential and is exactly what step 2.1 uses. An infinite family of positive radii need have no positive lower bound, and the minimum of an infinite set of reals need not exist at all (Every nonempty finite set of reals has a maximum and a minimum is stated for finite sets for that reason). The intersection of the balls $B (x, 1 / n)$ over all $n \geq 1$ is a standard example of an intersection of open sets that need not be open.
The empty intersection is not covered and does not need to be. Claim 3 is stated for $n \geq 1$ ; the conventional value of an empty intersection is $X$ , which is open anyway (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).
Claim 4 is not the statement that $\overset{ˉ}{B} (x, r)$ is the closure of $B (x, r)$ , which is false in general (FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$ ). All that is proved here is that the closed ball is a closed set.

Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources