Let *Y* be a subset of a topological space *X*. In the previous article, we defined the closure of *Y* as the smallest closed subset of *X* containing *Y*. Dually, we shall now define the interior of *Y* to be the largest open subset contained in it. The construction is similar.

- Let Σ be the collection of all open contained in
*Y*. - Since we see that Σ is not empty.
- Hence, it makes sense to define the
**interior**of*Y*as the union of all*U*in Σ.

Thus, As before, the interior satisfies the following.

- : since a union of open subsets is open, so int(
*Y*) must be open;

- if then since int(
*Y*) is a union of many sets including*U*, we must have

This justifies describing int(*Y*) as the largest open subset contained in *Y*.

As an immediate property:

Basic Fact. If are subsets of X, then

Now int(*Y*) is an open subset contained in *Y* and hence *Z*. Thus, int(*Y*) must be contained in int(*Z*).

**Examples**

- If
*Y*is open in*X*, then int(*Y*) =*Y*. - Take the half-open interval The interior is (0, 1).
- The the set of rationals
*Y*=**Q**in*X*=**R**. The interior is empty. - Suppose
*Y*is the set of even positive integers in**N***, together with ∞. What is the interior of*Y*? [ Answer: empty set. ]

Suppose *X* is a metric space. Now the open ball *N*(*a*, ε) is always open, but the interior of the closed ball *N*(*a*, ε)* is not the open ball *N*(*a*, ε) in general. For example, suppose *X* is the subspace [-1, 1] of **R**. Then the closed ball *Y* := *N*(0, 1)* is already open, so int(*Y*) = *Y*, whereas the open ball *N*(0, 1) is a proper subset of *Y*.

## Properties of Interior

One can relate the interior of *Y* with the closure of the complement of *Y*.

Proposition 1. For any subset , we have:

**Proof**.

Now int(*Y*) is the union of all open *U* contained in *Y*. Thus its complement is the intersection of all closed *C* which contains *Y*. This gives which is exactly what we desired to prove. ♦

Proposition 2. If are subsets, then

**Proof**.

We’ll turn this into a relation of closure by invoking proposition 1. Taking the complement turns the LHS into:

where the third equality followed from proposition 2 of the previous article. ♦

By induction, this can be generalised to finite intersections:

for arbitrary subsets

This does not hold for infinite intersections. For example, if then the intersection On the other hand,

It also doesn’t work for finite union. E.g. let *Y* = **Q** and *Z* = **R**–**Q**, in the real line *X* = **R**. Then int(*Y*) and int(*Z*) are both empty so But

However, we do have:

and

for any collection of subsets

- For the first inclusion: since for each
*i*, we have we get Since this works for each*i*, we’re done. - For the second inclusion: for each
*i*, we have so this gives Since this holds for each*i*, we’re done.

## Further Properties

Unfortunately, the interior of *Y* in a smaller subspace cannot be deduced from its interior in a bigger space.

Proposition 3. Let be subsets of a topological space X. Then:In general equality doesn’t hold.

**Proof**.

Since the RHS is an open subset of *Z* and is contained in *Y*, it must be contained in the LHS.

To see that equality doesn’t hold in general, let *X *= **R**, *Z *= **Q** and *Y* = **Q** ∩ (0, 1). Then since (0, 1) is open in **R**, *Y* is open in *Z*. Hence, On the other hand, so the RHS is empty. ♦

Thankfully, product still works:

Proposition 4. If and are subsets of topological spaces, thenwhere the LHS interior is taken in

**Proof**.

Since is an open subset of and is contained in we see that the RHS is contained in the LHS.

Conversely, if then there’s a basic open subset *U*×*V* such that Then and so and So the LHS is contained in the RHS. ♦

Finally, since we define the **boundary** of *Y* to be the difference

bd(*Y*) := cl(*Y*) – int(*Y*).

Note that since bd(*Y*) is the intersection of cl(*Y*) and the complement of int(*Y*), both closed, bd(*Y*) is closed too.

**Note**: some people also call cl(*Y*)-int(*Y*) the **frontier** of *Y*, since the term “boundary” has a different meaning in algebraic topology. We opt not to use that term, since in practice there’s little cause for confusion.

## Comparing Interior and Closure

The following table summarises the contents of this and the previous article. Some of these relations are equivalent by virtue of the fact that *X*-int(*Y*) = cl(*X*–*Y*).

Closure of Y |
Interior of Y |

If then | If then |

Finite union: | Finite intersection: |

Arbitrary union: | Arbitrary union: |

Arbitrary intersection: | Arbitrary intersection: |

If , then | If then |

For subsets we have | For subsets we have |

**Exercises**: deduce the corresponding entries for boundary bd(*Y*).

**Answers**:

- If
*Y*is contained in*Z*, it’s not always true that bd(*Y*) is contained in bd(*Z*). Indeed, take*Y*= (0, 1) ∩**Q**and*Z*= (0, 1). Then the boundary of*Y*is cl(*Y*) – int(*Y*) = [0, 1] – (empty set) = [0, 1], but the boundary of*Z*is {0, 1}. Surprised? - For finite union, bd(union of
*Y*,*Z*) is contained in the union of bd(*Y*) and bd(*Z*). - For finite intersection, bd(
*Y*∩*Z*) is contained in bd(*Y*) ∩ bd(*Z*). - For arbitrary union, things get hairy. E.g. if
*Y*= [0, (_{n}*n*-1)/*n*] for*n*=2, 3, …, then union of bd(*Y*) is {0, 1/2, 2/3, 3/4, … }, while bd(union of_{n}*Y*) = {0, 1}. Neither is contained in the other._{n} - Same for arbitrary intersection.
- bd
_{Z}(*Y*) is contained in*Z*∩ bd_{X}(*Y*). - bd(
*Y*_{1}×*Y*_{2}) is the union of bd(*Y*_{1}) × cl(*Y*_{2}) and cl(*Y*_{1}) × bd(*Y*_{2}).

Hello, thank you very much for this article. It is not clear to me what LHS and RHS mean

LHS and RHS refer to left-hand side and right-hand side respectively. E.g. in the equation , the RHS is 1.