## Topology: Closure

Suppose Y is a subset of a topological space X. We define cl(Y) to be the “smallest” closed subset containing Y. Its formal definition is as follows.

• Let Σ be the collection of all closed subsets $C\subseteq X$ containing Y.
• Note that $X\in \Sigma$, so Σ is not empty.
• Thus, we can take define $\text{cl}(Y) := \cap_{C\in\Sigma} C.$

This is called the closure of Y in X. Note that the notation cl(Y) does not indicate the ambient space X. If there’s any possibility of confusion, we denote it by clX(Y) instead. The closure satisfies the following properties:

• $\text{cl}(Y) \in \Sigma$: since an intersection of closed subsets is still closed;
• if $C\in \Sigma,$ then since cl(Y) is the intersection of many sets, including C, we must have $\text{cl}(Y)\subseteq C.$

This justifies our calling cl(Y) the smallest closed subset containing Y.

As an immediate property, we have:

Basic Fact. If $Y\subseteq Z$ are subsets of X, then $\text{cl}(Y)\subseteq \text{cl}(Z).$

Indeed, cl(Z) is a closed subset of X containing Z, and hence Y. Thus, it must contain cl(Y).

Examples

1. If Y is closed in X, then cl(Y) = Y.
2. Take the half-open interval $Y = [0, 1)\subseteq \mathbf{R}.$ The closure is [0, 1].
3. Take the set of rationals YQ in XR. The closure is the whole real line.
4. Suppose Y = {3, 5, 7} in N*. What is the closure of Y (see here for the definition of N*)? [ Answer: {1, 2, 3, 4, 5, 6, 7}. ]

Warning: in a metric space X, the closed ball $N(a, \epsilon)^* := \{x\in X: d(x, a) \le \epsilon\}$ is a closed subset since the complement (which contains all x such that d(xa) > ε) is open in X.

However, the closure of the open ball $N(a, \epsilon)$ is not the closed ball $N(a, \epsilon)^*$ in general. For a rather pathological example, consider $X = [-1, +1] \cup \{2\}$ and take the open ball Y = N(0, 2). Then Y = [-1, +1] is already closed in X so cl(Y) = Y, while the closed ball N(0, 2)* = X ≠ cl(Y).

## Properties of Closure

The closure of can be characterised as follows.

Proposition 1. We have $\text{cl}(Y) = Y\cup Y^{acc}$ where Yacc is the collection of all points of accumulation of Y.

Proof.

Let’s first prove that $C:=Y\cup Y^{acc}$ is closed in X.

• If $x\in X-C,$ then in particular x is not a point of accumulation of Y, so there exists an open subset $U\subseteq X$ containing x such that U ∩ Y contains at most x. But x is not in Y, so $U\cap Y=\emptyset,$ i.e. $U\subseteq X-Y.$
• Next we wish to show $U\subseteq X-Y^{acc}.$ Suppose on the contrary $y\in U\cap Y^{acc},$ so in particular y is a point of accumulation of Y. Since U is an open subset of X containing y, we have U ∩ Y containing a point z ≠ y, which contradicts what we just proved. Thus $U\subseteq X - (Y\cup Y^{acc}) = X-C,$ so XC is a union of open subsets and is thus open.

To complete the proof, since C is a closed subset containing Y, it also contains cl(Y). On the other hand, since cl(Y) is closed, $\text{cl}(Y) \supseteq \text{cl}(Y)^{acc} \supseteq Y^{acc}.$ This gives $\text{cl}(Y) \supseteq Y\cup Y^{acc} = C,$ which completes the proof. ♦

Proposition 2. If $Y, Z\subseteq X$ are subsets, then

$\text{cl}(Y\cup Z) = \text{cl}(Y) \cup \text{cl}(Z).$

Proof.

Since the RHS is closed and contains $Y\cup Z$, it must contain the LHS also.

Conversely, since $Y\subseteq Y\cup Z,$ we have $\text{cl}(Y) \subseteq \text{cl}(Y\cup Z).$ Similarly, $\text{cl}(Z) \subseteq\text{cl}(Y\cup Z)$ so the LHS contains the RHS. ♦

## Unions and Intersections of Closures

By induction, one can prove that:

$\text{cl}(Y_1 \cup Y_2 \cup \ldots \cup Y_n) = \text{cl}(Y_1) \cup \text{cl}(Y_2) \cup \ldots \cup \text{cl}(Y_n)$

for any subsets $Y_1, Y_2, \ldots, Y_n \subseteq X.$ However, this is as far as we go, as we cannot extend this to the union of infinitely many sets. E.g. if $Y_n = [\frac 1 n, 1]$ in X=R, then since each $Y_n\subseteq X$ is closed, we have $\cup_n\text{cl}(Y_n) = \cup_n Y_n=(0, 1]$ which isn’t closed, so it cannot possibly be $\text{cl}(\cup_n Y_n).$

Also, the result is not true for intersection. E.g. let YQ and ZRQ be subsets of the real line R. Then Y ∩ Z is empty, and so is cl(Y ∩ Z). On the other hand, cl(Y) ∩ cl(Z) = R ∩ RR. So we have cl(Y) ∩ cl(Z) ≠ cl(Y ∩ Z).

That being said, we do at least have a one-sided inclusions:

$\text{cl}(\cap_i Y_i) \subseteq \cap_i \text{cl}(Y_i), \quad \text{cl}(\cup_i Y_i) \supseteq \cup_i \text{cl}(Y_i)$

for any collection of subsets $Y_i\subseteq X.$

• For the first inclusion: since $\cap_i Y_i \subseteq Y_i$ for each i, we have $\text{cl}(\cap_i Y_i)\subseteq \text{cl}(Y_i).$ Since this holds for every i, we get the desired inclusion.
• For the second inclusion: since $Y_i\subseteq \cup_i Y_i$ for each i, we have $\text{cl}(Y_i) \subseteq \text{cl}(\cup_i Y_i).$ Since this holds for each i, we’re done. ♦

## Further Properties

Next, the closure of Y in a smaller subspace can be deduced from the closure in a bigger space.

Proposition 3. If $Y\subseteq Z \subseteq X$ are subsets of a topological space X, then the closure of Y in Z is the intersection of Z with the closure of Y in X.

$\text{cl}_Z(Y) = Z\cap \text{cl}_X(Y).$

Proof

Since the RHS is a closed subset of Z containing Y, it must contain the LHS.

Conversely, since LHS is closed in Z it must be of the form C ∩ Z for some closed subset C of X. Then C is a closed subset of X containing Y and thus must contain clX(Y). So C ∩ Z must contain the RHS. ♦

Finally, closure also respects the product.

Proposition 4. If $Y_1 \subseteq X_1$ and $Y_2\subseteq X_2$ are subsets of topological spaces, then:

$\text{cl}(Y_1\times Y_2) = \text{cl}(Y_1) \times \text{cl}(Y_2)$

where the LHS closure is taken in $X_1\times X_2.$

Proof.

Since the RHS is closed in $X_1 \times X_2,$ it must contain the LHS. For the reverse inclusion, we prove by contradiction by picking an element (xy) outside the LHS, cl(Y1 × Y2).

• In particular, $(x, y)\not\in (Y_1\times Y_2)^{acc},$ so there is an open subset containing (xy) which doesn’t intersect Y1 × Y2.
• This open subset must contain a basic open set of the form U × V which contains (xy).
• Note that we still have $(U\times V)\cap (Y_1\times Y_2) = \emptyset$ and so $U\cap Y_1 = \emptyset$ or $V\cap Y_2 = \emptyset.$
• Assume the former; then $x\not\in Y_1^{acc}.$ Since $x\in U,$ we also have $x\not\in Y_1,$ thus giving $x\not\in Y_1\cup Y_1^{acc} = \text{cl}(Y_1).$

So (xy) is outside the RHS also, and we’re done. ♦

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