Topology/Points in Sets

Some Important Constructions

Let $X$ be a topological space and $A$ be any subset of $X$.

Closure

• A point $x$ is called a point of closure of $A$ if every neighborhood of $x$ contains at least one element of $A$. In other words, for all neighborhoods $U$ of $x$, $U\cap A\ne \mathrm{\varnothing }$.
• The closure of $A$ is the set of all points of closure of $A$. It is equivalent to the intersection of all closed sets that contain $A$ as a subset, denoted $\mathrm{C}\mathrm{l}(A)$ (some authors use $\overline{A}$). Alternatively, it is the set $A$ together with all its limit points (defined below). The closure has the nice property of being the smallest closed set containing $A$. All neighborhoods of each point in the closure intersects $A$.

Interior

• A point $x$ is an internal point of $A$ if there is an open subset of $A$ containing $x$.
• The interior of $A$ is the union of all open sets contained inside $A$, denoted $\mathrm{I}\mathrm{n}\mathrm{t}(A)$ (some authors use $A^{\circ }$). The interior has the nice property of being the largest open set contained inside $A$. Every point in the interior has a neighborhood contained inside $A$. It is equivalent to the set of all interior points of $A$.

Note that an open set is equal to its interior.

Exterior

• Define the exterior of $A$ to be the union of all open sets contained inside the complement of $A$, denoted $\mathrm{I}\mathrm{n}\mathrm{t}(X\setminus A)$. It is the largest open set inside $X\setminus A$. Every point in the exterior has a neighborhood contained inside $X\setminus A$.

Boundary

• Define the boundary of $A$ to be the closure of $A$ excluding its interior, or $\mathrm{C}\mathrm{l}(A)\setminus \mathrm{I}\mathrm{n}\mathrm{t}(A)$. It is denoted $\mathrm{B}\mathrm{d}(A)$ (some authors prefer $\partial A$). The boundary is also called the frontier. It is always closed since it is the intersection of the closed set $\mathrm{C}\mathrm{l}(A)$ and the closed set $X\setminus \mathrm{I}\mathrm{n}\mathrm{t}(A)$. It can be proved that $A$ is closed if it contains all its boundary, and is open if it contains none of its boundary. Every neighborhood of each point in the boundary intersects both $A$ and $X\setminus A$. All boundary points of a set $A$ are obviously points of contact of $A$.

Limit Points

• A point $x$ is called a limit point of $A$ if every neighborhood of $x$ intersects $A$ in at least one point other than $x$. In other words, for every neighborhood $U$ of $x$, $(U\setminus \{x\})\cap A\ne \mathrm{\varnothing }$. All limit points of $A$ are obviously points of closure of $A$.

Isolated Points

• A point $x$ of $A$ is an isolated point of $A$ if it has a neighborhood which does not contain any other points of $A$. This is equivalent to saying that $\{x\}$ is an open set in the topological space $A$ (considered as a subspace of $X$).

Density

Definition: $A$ is called dense (or dense in $X$) if every point in $X$ either belongs to $A$ or is a limit point of $A$. Informally, every point of $X$ is either in $A$ or arbitrarily close to a member of $A$. For instance, the rational numbers are dense in the real numbers because every real number is either a rational number or has a rational number arbitrarily close to it.

Equivalently: $A$ is dense if the closure of $A$ is $X$.

Definition: $A$ is nowhere dense (or nowhere dense in $X$) if the closure of $A$ has an empty interior. That is, the closure of $A$ contains no non-empty open sets. Informally, it is a set whose points are not tightly clustered anywhere. For instance, the set of integers is nowhere dense in the set of real numbers. Note that the order of operations matters: the set of rational numbers has an interior with empty closure, but it is not nowhere dense; in fact it is dense in the real numbers.

Definition: A G_σ set is a subset of a topological space that is a countable intersection of open sets.

Definition: An F_σ set is a countable union of closed sets.

Theorem

(Hausdorff Criterion) Suppose X has 2 topologies, r₁ and r₂. For each $x\in X$, let B¹_x be a neighbourhood base for x in topology r₁ and B²_x be a neighbourhood base for x in topology r₂. Then, $r_1\subseteq r_2$ if and only if at each $x\in X$, if $(B^1\in B_x^1)(\mathrm{\exists }(B^2\in B_x^2)(B^2\subseteq B^1).$

Theorem

In any topological space, the boundary of an open set is closed and nowhere dense.

Proof:
Let A be an open set in a topological space X. Since A is open, int(A) = A. Thus, $\partial A$ ( or the boundary of A) = $\overline{A}/int(A)$. Note that $\overline{A}/A=\overline{A}\cap A^c$. The complement of an open set is closed, and the closure of any set is closed. Thus, $\overline{A}\cap A^c$ is an intersection of closed sets and is itself closed. A subset of a topological space is nowhere dense if and only if the interior of its closure is empty. So, proceeding in consideration of the boundary of A.

The interior of the closure of the boundary of A is equal to the interior of the boundary of A.

Thus, it is equal to $int(\overline{A}\cap A^c)$.

Which is also equal to $int(\overline{A})\cap int(A^c)$.

And, $int(A^c)={\overline{[(A^c{)}^c]}}^c$. So, the interior of the closure of the boundary of A = $int(\overline{A})\cap int\overline{(A^c)}$., and as such, the boundary of A is nowhere dense.

Types of Spaces

We can also categorize spaces based on what kinds of points they have.

Perfect Spaces

• If a space contains no isolated points, then the space is a perfect space.

Some Basic Results

• For every set $A$; $A\subseteq \mathrm{C}\mathrm{l}(A)$ and $\mathrm{I}\mathrm{n}\mathrm{t}(A)\subseteq A$
  Proof:
  Let $x\in A$. If a closed set $\alpha \supseteq A$, then $x\in \alpha$. As $\mathrm{C}\mathrm{l}(A)=\bigcap _{\alpha \subset X}\alpha$ for closed $\alpha$; we have $x\in \mathrm{C}\mathrm{l}(A)$. $x\in A$ being arbitrary, $A\subseteq \mathrm{C}\mathrm{l}(A)$
  Let $U\subseteq A$ be open. Thus, $x\in A\mathrm{\forall }x\in U$. As $\mathrm{I}\mathrm{n}\mathrm{t}(A)=\bigcup _{U\subseteq A}U$ for open $U$; we have $x\in \mathrm{I}\mathrm{n}\mathrm{t}(A)\mathrm{\forall }x\in U$. $U\subseteq A$ being arbitrary, we have $\mathrm{i}\mathrm{n}\mathrm{t}(A)\subseteq A$

• A set $A$ is open if and only if $\mathrm{I}\mathrm{n}\mathrm{t}(A)=A$.
  Proof:
  ($⟹$)
  $A$ is open and $A\subseteq A$. Hence, $A\subseteq \mathrm{I}\mathrm{n}\mathrm{t}(A)$. But we know that $\mathrm{I}\mathrm{n}\mathrm{t}(A)\subseteq A$ and hence $\mathrm{I}\mathrm{n}\mathrm{t}(A)=A$
  ($⟸$)
  As $\mathrm{I}\mathrm{n}\mathrm{t}(A)$ is a union of open sets, it is open (from definition of open set). Hence $A=\mathrm{I}\mathrm{n}\mathrm{t}(A)$ is also open.

• A set $A$ is closed if and only if $\mathrm{C}\mathrm{l}(A)=A.$
  Proof:
  Observe that the complement of $\mathrm{C}\mathrm{l}(A)$ satisfies $X\setminus \mathrm{C}\mathrm{l}(A)=\mathrm{I}\mathrm{n}\mathrm{t}(X\setminus A)$. Hence, the required result is equivalent to the statement "$X\setminus A$ is open if and only if $\mathrm{I}\mathrm{n}\mathrm{t}(X\setminus A)=X\setminus A$". $A$ is closed implies that $X\setminus A$ is open, and hence we can use the previous property.

• The closure $\mathrm{C}\mathrm{l}(A)$ of a set $A$ is closed
  Proof:
  Let $\alpha$ be a closed set such that $A\subseteq \alpha$. Now, $\mathrm{C}\mathrm{l}(A)=\bigcap _{\alpha \subset X}\alpha$ for closed $\alpha$. We know that the intersection of any collection of closed sets is closed, and hence $\mathrm{C}\mathrm{l}(A)$ is closed.

Exercises

1. Prove the following identities for subsets $A,B$ of a topological space $X$:
   • $\mathrm{C}\mathrm{l}(A\cup B)=\mathrm{C}\mathrm{l}(A)\cup \mathrm{C}\mathrm{l}(B)$
   • $\mathrm{C}\mathrm{l}(A\cap B)\subseteq \mathrm{C}\mathrm{l}(A)\cap \mathrm{C}\mathrm{l}(B)$
   • $\mathrm{I}\mathrm{n}\mathrm{t}(A)\cup \mathrm{I}\mathrm{n}\mathrm{t}(B)\subseteq \mathrm{I}\mathrm{n}\mathrm{t}(A\cup B)$
   • $\mathrm{I}\mathrm{n}\mathrm{t}(A\cap B)=\mathrm{I}\mathrm{n}\mathrm{t}(A)\cap \mathrm{I}\mathrm{n}\mathrm{t}(B)$
2. Show that the following identities need not hold (i.e. give an example of a topological space and sets $A$ and $B$ for which they fail):
   • $\mathrm{C}\mathrm{l}(A\cap B)=\mathrm{C}\mathrm{l}(A)\cap \mathrm{C}\mathrm{l}(B)$
   • $\mathrm{I}\mathrm{n}\mathrm{t}(A)\cup \mathrm{I}\mathrm{n}\mathrm{t}(B)=\mathrm{I}\mathrm{n}\mathrm{t}(A\cup B)$

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