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 $$x$$, $$U \cap A \neq \emptyset$$.
 * 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 $$\mathrm{Cl}(A)$$ (some authors use $$\bar{A}$$). 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 $$\mathrm{Int}(A)$$ (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 $$\mathrm{Int} (X \setminus A)$$. 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 $$\mathrm{Cl}(A)\setminus\mathrm{Int}(A)$$. It is denoted $$\mathrm{Bd}(A)$$ (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 $$\mathrm{Cl}(A)$$ and the closed set $$X\setminus\mathrm{Int}(A)$$. 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 $$x$$, $$(U\setminus\{x\}) \cap A \neq \emptyset$$. 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 Gσ set is a subset of a topological space that is a countable intersection of open sets.

Definition: An Fσ 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 B1x be a neighbourhood base for x in topology r1 and B2x 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 $$ (B^1 \in B^1_x)(\exists (B^2 \in B^2_x)( B^2 \subseteq B^1).$$

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) = $$\bar{A}/  int (A) $$. Note that $$ \bar A/A = \bar A \cap A^c $$. The complement of an open set is closed, and the closure of any set is closed. Thus, $$\bar A \cap A^c $$ 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 $$int (\bar{A} \cap A^c)$$.
 * Which is also equal to $$int (\bar{A}) \cap int (A^c)$$.

And, $$int (A^c) = \bar{[(A^c)^c]}^c$$. So, the interior of the closure of the boundary of A = $$int (\bar{A}) \cap int \bar{(A^c)}$$., 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
Proof: Let $$x\in A$$. If a closed set $$\alpha\supseteq A$$, then $$x\in\alpha$$. As $$\mathrm{Cl}(A)=\displaystyle\bigcap_{\alpha\subset X} \alpha$$ for closed $$\alpha$$; we have $$x\in\mathrm{Cl}(A)$$. $$x\in A$$ being arbitrary, $$A\subseteq \mathrm{Cl}(A)$$ Let $$U\subseteq A$$ be open. Thus, $$x\in A\forall x\in U$$. As $$\mathrm{Int}(A)=\displaystyle\bigcup_{U\subseteq A}U$$ for open $$U$$; we have $$x\in \mathrm{Int}(A)\forall x\in U$$. $$U\subseteq A$$ being arbitrary, we have $$\mathrm{int}(A)\subseteq A$$
 * For every set $$A$$; $$A\subseteq \mathrm{Cl}(A)$$ and $$\mathrm{Int}(A)\subseteq A$$

Proof: ($$\Longrightarrow$$) $$A$$ is open and $$A\subseteq A$$. Hence, $$A\subseteq\mathrm{Int}(A)$$. But we know that $$\mathrm{Int}(A)\subseteq A$$ and hence $$\mathrm{Int}(A)=A$$ ($$\Longleftarrow$$) As $$\mathrm{Int}(A)$$ is a union of open sets, it is open (from definition of open set). Hence $$A=\mathrm{Int}(A)$$ is also open.
 * A set $$A$$ is open if and only if $$\mathrm{Int}(A)=A$$.

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

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

Exercises

 * 1) Prove the following identities for subsets $$A,B$$ of a topological space $$X$$:
 * 2) *$$\mathrm{Cl}(A\cup B)=\mathrm{Cl}(A)\cup\mathrm{Cl}(B)$$
 * 3) *$$\mathrm{Cl}(A\cap B)\subseteq\mathrm{Cl}(A)\cap\mathrm{Cl}(B)$$
 * 4) *$$\mathrm{Int}(A)\cup\mathrm{Int}(B)\subseteq\mathrm{Int}(A\cup B)$$
 * 5) *$$\mathrm{Int}(A\cap B)=\mathrm{Int}(A)\cap\mathrm{Int}(B)$$
 * 6) 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):
 * 7) *$$\mathrm{Cl}(A\cap B)=\mathrm{Cl}(A)\cap\mathrm{Cl}(B)$$
 * 8) *$$\mathrm{Int}(A)\cup\mathrm{Int}(B)=\mathrm{Int}(A\cup B)$$