Union and Intersection of Closed Sets¶

Proposition¶

For a metric space $(X,d)$ and a collection of closed subsets $(E_i)_{i=1}^\infty$ of $X$, the countable intersection $\displaystyle\bigcap_{i=1}^\infty E_i$ and the finite union $\displaystyle \bigcup_{i=1}^k E_i$ are also closed.

Interpretation¶

Closed sets are preserved under finite unions and arbitrary intersections. Infinite unions can break closedness, because a limit of points from infinitely many sets need not lie in any one of them.

Each $F_n=[\tfrac{1}{n},1]$ is closed, and their union over all $n$ is $(0,1]$. The left endpoints $\tfrac{1}{n}$ all lie in the union and converge to $0$, but $0$ lies in no $F_n$, so the union misses a limit of its own points and is not closed. Finitely many $F_n$ never do this: $F_1\cup\dots\cup F_k=[\tfrac{1}{k},1]$ is closed.

Both statements can be read through sequences, since a set is closed if and only if it contains the limits of its convergent sequences. A convergent sequence in $\bigcap_i E_i$ lies in every $E_i$, so its limit does too. A sequence in $E_1\cup\dots\cup E_k$ has infinitely many terms in one $E_j$; this subsequence has the same limit, which therefore lies in $E_j$. With infinitely many sets no single set need contain a whole subsequence, as in the figure.

The proposition is also the mirror image of the union and intersection of open sets: taking complements turns intersections into unions and closed sets into open sets. Arbitrary intersections are why the intersection of all closed sets containing $E$ is itself closed; it is the closure of $E$.

A common slip is to read the figure as "infinite unions of closed sets are never closed". They may or may not be: the union of all $[n,n+1]$ with $n\in\mathbb{Z}$ is $\mathbb{R}$, which is closed. And $(0,1]$ is not open either.

Key Results¶

• Both $X$ and $\varnothing$ are always closed (and also open).
• In fact any intersection of closed sets, not just countably many, is closed. This holds for uncountable families as well.

Examples¶

• In $\mathbb{R}$, $\displaystyle \bigcap_{n=1}^\infty [-1/n, 1/n] = \{0\}$ is closed. Shrinking closed intervals collapse to a single closed point.
• In $\mathbb{R}$, $\displaystyle \bigcup_{n=1}^\infty [0, 1-1/n] = [0,1)$ is not closed. It is missing the endpoint $1$.

Prerequisites¶

• Open and Closed Sets in a Metric Space
• Countable and Uncountable Sets
• Union of Sets

Related Concepts¶

• Union and Intersection of Open Sets

Sources¶

• Math III: Analysis · Lecture 6

Symbols: element of, infinity, integers, real numbers, union

Last updated: 20 October 2026 03:02