## Topology: Interior

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 $U\subseteq X$ contained in Y.
• Since $\emptyset \in\Sigma,$ 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, $\text{int}(Y) = \cup_{U\in\Sigma} U.$ As before, the interior satisfies the following.

• $\text{int}(Y)\in\Sigma$ : since a union of open subsets is open, so int(Y) must be open;
• if $U\in\Sigma,$ then since int(Y) is a union of many sets including U, we must have $U\subseteq \text{int}(Y).$

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

As an immediate property:

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

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

Examples

1. If Y is open in X, then int(Y) = Y.
2. Take the half-open interval $Y=[0, 1)\subseteq \mathbf{R}.$ The interior is (0, 1).
3. The the set of rationals YQ in XR. The interior is empty.
4. 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 $Y\subseteq X$, we have:

$\text{int}(Y) = X - \text{cl}(X-Y).$

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 $X-\text{int}(Y) = \text{cl}(X-Y),$ which is exactly what we desired to prove. ♦

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

$\text{int}(Y\cap Z) = \text{int}(Y)\cap\text{int}(Z).$

Proof.

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

\begin{aligned}X-\text{int}(Y\cap Z) &=\text{cl}(X - (Y\cap Z)) = \text{cl}((X-Y)\cup(X-Z))= \text{cl}(X-Y)\cup\text{cl}(X-Z)\\&= (X-\text{int}(Y))\cup (X-\text{int}(Z)) = X - (\text{int}(Y)\cap\text{int}(Z)),\end{aligned}

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

By induction, this can be generalised to finite intersections:

$\text{int}(Y_1) \cap \text{int}(Y_2) \cap\ldots\cap \text{int}(Y_n) = \text{int}(Y_1\cap Y_2 \cap \ldots \cap Y_n)$

for arbitrary subsets $Y_1, Y_2, \ldots, Y_n\subseteq X.$

This does not hold for infinite intersections. For example, if $Y_n = [-\frac 1 n, +\frac 1 n],$ then the intersection $\cap_n \text{int}(Y_n) = \cap_n (-\frac 1 n, +\frac 1 n) = \{0\}.$ On the other hand, $\text{int}(\cap_n Y_n) = \text{int}(\{0\}) = \emptyset.$

It also doesn’t work for finite union. E.g. let YQ and ZRQ, in the real line XR. Then int(Y) and int(Z) are both empty so $\text{int}(Y)\cup\text{int}(Z) = \emptyset.$ But $\text{int}(Y\cup Z) = \text{int}(\mathbf{R}) = \mathbf{R}.$

However, we do have:

$\text{int}(\cup_i Y_i) \supseteq \cup_i \text{int}(Y_i)\$ and $\text{int}(\cap Y_i) \subseteq \cap_i \text{int}(Y_i)$

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

• For the first inclusion: since for each i, we have $Y_i\subseteq \cup_i Y_i,$ we get $\text{int}(Y_i)\subseteq \text{int}(\cup_i Y_i).$ Since this works for each i, we’re done.
• For the second inclusion: for each i, we have $\cap_i Y_i \subseteq Y_i$ so this gives $\text{int}(\cap_i Y_i) \subseteq \text{int}(Y_i).$ 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 $Y\subseteq Z\subseteq X$ be subsets of a topological space X. Then:

$\text{int}_Z(Y) \supseteq Z\cap\text{int}_X(Y).$

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 RQ and YQ ∩ (0, 1). Then since (0, 1) is open in RY is open in Z. Hence, $\text{int}_Z(Y) = Y.$ On the other hand, $\text{int}_X(Y) = \emptyset,$ so the RHS is empty. ♦

Thankfully, product still works:

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

$\text{int}(Y_1 \times Y_2) = \text{int}(Y_1)\times \text{int}(Y_2),$

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

Proof.

Since $\text{int}(Y_1)\times \text{int}(Y_2)$ is an open subset of $X_1\times X_2$ and is contained in $Y_1\times Y_2,$ we see that the RHS is contained in the LHS.

Conversely, if $(x,y)\in \text{int}(Y_1\times Y_2)$ then there’s a basic open subset U×V such that $(x,y)\in U\times V\subseteq \text{int}(Y_1 \times Y_2).$ Then $U\subseteq Y_1$ and $V\subseteq Y_2$ so $x\in \text{int}(Y_1)$ and $y\in \text{int}(Y_2).$ So the LHS is contained in the RHS. ♦

Finally, since $\text{int}(Y) \subseteq Y\subseteq \text{cl}(Y)$ 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(XY).

 Closure of Y Interior of Y If $Y\subseteq Z,$ then $\text{cl}(Y)\subseteq \text{cl}(Z).$ If $Y\subseteq Z,$ then $\text{int}(Y) \subseteq \text{int}(Z).$ Finite union: $\text{cl}(Y\cup Z) = \text{cl}(Y) \cup \text{cl}(Z).$ Finite intersection: $\text{int}(Y \cap Z) = \text{int}(Y) \cap \text{int}(Z).$ Arbitrary union: $\text{cl}(\cup_i Y_i) \supseteq \cup_i \text{cl}(Y_i).$ Arbitrary union: $\text{int}(\cup_i Y_i) \supseteq \cup_i \text{int}(Y_i).$ Arbitrary intersection: $\text{cl}(\cap_i Y_i) \subseteq \cap_i \text{cl}(Y_i).$ Arbitrary intersection: $\text{int}(\cap_i Y_i) \subseteq \cap_i\text{int}(Y_i).$ If $Y\subseteq Z\subseteq X$, then $\text{cl}_Z(Y) = Z\cap \text{cl}_X(Y).$ If $Y\subseteq Z\subseteq X,$ then $\text{int}_Z(Y) \supseteq Z\cap \text{int}_X(Y).$ For subsets $Y_1\subseteq X_1, Y_2\subseteq X_2,$ we have $\text{cl}(Y_1\times Y_2) = \text{cl}(Y_1)\times \text{cl}(Y_2).$ For subsets $Y_1\subseteq X_1, Y_2\subseteq X_2,$ we have $\text{int}(Y_1\times Y_2) = \text{int}(Y_1)\times \text{int}(Y_2).$

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