Topological Spaces

1. Topological Space

Definition. Topology and Topological Space, and Open Sets.

A pair $(X,\Sigma)$ is called a topological space if $X$, as a set, is equipped with a collection $\Sigma$ of subsets (called a topology) of $X$ that satisfies:

  1. $\emptyset\in\Sigma$, $X\in\Sigma$

  2. Arbitraty union is in $\Sigma$. i.e.

    $$\{U_\alpha\}_{\alpha\in\Lambda}\subset\Sigma\implies \bigcup_{\alpha\in\Lambda}U_{\alpha}\in\Sigma $$
  3. Finite intersection is in $\Sigma$, i.e.

    $$\{U_i\}_{i=1}^n\subset\Sigma\implies \bigcap_{i=1}^nU_{i}\in\Sigma $$

In this case, every $U$ in $\Sigma$ is called an open set.

Example. $(X, d)$ is a metric space,

  1. $(X, \{\emptyset, X\})$ is called the trivial space.
  2. $(X, 2^X)$ is called the discrete space.

Definition. Subspace Topology and Subspace

$(X, \Sigma)$ is a topological space with $Y\subset X$. The the collection

$$\Sigma_Y:=\{Y\cap U\mid U\in \Sigma\} $$

is called a subspace topology of $\Sigma$, and $(Y,\Sigma_Y)$ is called a subspace of $(X, \Sigma)$ correspondingly.

Definition. Closedness

$(X, \Sigma)$ is a topological space, $S$ is called closed in $X$ if $X\setminus S\in \Sigma$

Same in metric space, closedness is not contrary to openness.

Lemma. $(X, \Sigma)$ is a topological space, then the followings are true

  1. $\emptyset$ and $X$ are closed.
  2. arbitrary intersection of closed subsets are is closed.
  3. finite union of closed subsets is closed.

Lemma. $Y\subset X$ is a subspace of $X$, then $S$ is closed in $Y$ (with respect to the subspace topology) iff there exists a closed subset $E\subset X$ such that $S=E\cap Y$

Corollary, $Y\subset X$ is closed. $S\subset Y$ is closed in $Y$ iff $S$ is closed in $X$.

Definition. Interior Point Set, Closure

$\Sigma$ is a topology of $X$, the set of Interior Point of $S\subset X$ is defined as

$$S^\circ:=\bigcup\{U\in\Sigma\mid U\subset S\} $$

With $S$ defined, the closure of $S$ is defined as

$$\bar S:=\bigcap\{E\subset X\mid S\subset X, \text{E is closed}\} $$

By definition we have

$$S^\circ\subset S\subset \bar S $$

Theorem. Localization Property of Closure