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 containing
*Y*.

- Note that , so Σ is not empty.
- Thus, we can take define

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 cl_{X}(*Y*) instead. The closure satisfies the following properties:

- : since an intersection of closed subsets is still closed;
- if then since cl(
*Y*) is the intersection of many sets, including*C*, we must have

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

As an immediate property, we have:

Basic Fact. If are subsets of X, then

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

**Examples**

- If
*Y*is closed in*X*, then cl(*Y*) =*Y*. - Take the half-open interval The closure is [0, 1].
- Take the set of rationals
*Y*=**Q**in*X*=**R**. The closure is the whole real line. - 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** is a closed subset since the complement (which contains all *x* such that *d*(*x*, *a*) > ε) is open in *X*.

However, the closure of the open ball is *not* the closed ball in general. For a rather pathological example, consider 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 *Y *can be characterised as follows.

Proposition 1. We have where Y^{acc}is the collection of all points of accumulation of Y.

**Proof**.

Let’s first prove that is closed in *X*.

- If then in particular
*x*is not a point of accumulation of*Y*, so there exists an open subset containing*x*such that*U*∩*Y*contains at most*x*. But*x*is not in*Y*, so i.e. - Next we wish to show Suppose on the contrary 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 so*X*–*C*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, This gives which completes the proof. ♦

Proposition 2. If are subsets, then

**Proof**.

Since the RHS is closed and contains , it must contain the LHS also.

Conversely, since we have Similarly, so the LHS contains the RHS. ♦

## Unions and Intersections of Closures

By induction, one can prove that:

for any subsets However, this is as far as we go, as we cannot extend this to the union of infinitely many sets. E.g. if in *X*=**R**, then since each is closed, we have which isn’t closed, so it cannot possibly be

Also, the result is not true for intersection. E.g. let *Y* = **Q** and *Z* = **R**–**Q** 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** ∩ **R** = **R**. So we have cl(*Y*) ∩ cl(*Z*) ≠ cl(*Y* ∩ *Z*).

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

for any collection of subsets

- For the first inclusion: since for each
*i*, we have Since this holds for every*i*, we get the desired inclusion. - For the second inclusion: since for each
*i*, we have 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 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.

**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 cl_{X}(*Y*). So *C* ∩ *Z* must contain the RHS. ♦

Finally, closure also respects the product.

Proposition 4. If and are subsets of topological spaces, then:where the LHS closure is taken in

**Proof**.

Since the RHS is closed in it must contain the LHS. For the reverse inclusion, we prove by contradiction by picking an element (*x*, *y*) outside the LHS, cl(*Y*_{1} × *Y*_{2}).

- In particular, so there is an open subset containing (
*x*,*y*) which doesn’t intersect*Y*_{1}×*Y*_{2}. - This open subset must contain a basic open set of the form
*U*×*V*which contains (*x*,*y*). - Note that we still have and so or
- Assume the former; then Since we also have thus giving

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