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

If Y is open in X, then int(Y) = Y.
Take the half-open interval $Y=[0, 1)\subseteq \mathbf{R}.$ 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 $Y\subseteq X$ , 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 Y = Q and Z = R–Q, in the real line X = R. 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 X = R, Z = Q and Y = Q ∩ (0, 1). Then since (0, 1) is open in R, Y 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(X–Y).

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).

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_n = [0, (n-1)/n] for n=2, 3, …, then union of bd(Y_n) is {0, 1/2, 2/3, 3/4, … }, while bd(union of Y_n) = {0, 1}. Neither is contained in the other.
Same for arbitrary intersection.
bd_Z(Y) is contained in Z ∩ bd_X(Y).
bd(Y₁ × Y₂) is the union of bd(Y₁) × cl(Y₂) and cl(Y₁) × bd(Y₂).

2 Responses to Topology: Interior

Diego says:

January 26, 2021 at 1:06 am

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

- limsup says:
  
  January 27, 2021 at 3:02 pm
  
  LHS and RHS refer to left-hand side and right-hand side respectively. E.g. in the equation $\sin^2 x + \cos^2 x = 1$ , the RHS is 1.