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We need to revisit metric spaces since some definitions on topological spaces depend on it.
Let $X$ be a non-empty set and $d:X \times X \rightarrow \mathbb{R}$ a metric function. Then $(X, d)$ is a metric space.
Given a metric space $(X, d)$, a point $a \in X$ and a scalar $\delta \gt 0$, an open ball about $a$, denoted by $B(a, \delta)$ is a set of points $x \in X$ satisfying $d(a, x) \lt \delta$.
Given a metric space $(X, d)$, a subset $N$ of $X$ is a neighborhood of point $a \in X$ if there exists some $\delta \gt 0$ and $N$ contains $B(a, \delta)$.
Given a metric space $(X, d)$ and a subset $A \subseteq X$, a point $x \in X$ is limit point if every neighborhood of $x$ contains a point of $A$ different than $x$. Note that $x$ doesn’t need to be in $A$, nor that every point in $A$ is a limit point (e.g. in the set $\mathbb{N}$ no element is a limit point).
Example. If $A = \curly{1/x : x \in \mathbb{R}}$, then $0$ is a limit point of $A$ even though it’s not in $A$, since for any ball $B(0, \delta)$ contains $\delta > 0$ and $\delta \in A$.
Given a metric space $(X, d)$, a subset $O$ of $X$ is an open set if it’s a neighborhood of its points. That is, for every $a \in O$, there exists some $\delta$ such that $B(a, \delta) \subseteq O$.
Open sets satisfy the following properties:
A topological space is a generalization of a metric space. We can obtain a topological space from a metric space by discarding the metric function and instead working with the open sets which can be derived from the metric function.
Not every collection of open subsets of $X$ can be obtained via a metric function, so we can say that all metric spaces are a topological spaces but not the opposite.
Let $X$ be a non-empty set and $\tau$ a collection of subsets of $X$. These subsets satisfy the open set properties $O1$-$O4$ which we can state as follows:
Then $(X, \tau)$ is a topological space. $\tau$ is called the topology of set $X$. Members of $\tau$ are called open sets.
Given a topological space $(X, \tau)$ and a subset $N$ of $X$, $N$ is called a neighborhood of a point $a \in X$ if it contains an open set that contains $a$. That is, there exists $O \in \tau$ such that $a \in O$ and $O \subseteq N$.
Given a topological space $(X, \tau)$ and a subset $F$ of $X$, the complement of $F$, $C(F)$ is $X \setminus F$.
Given a topological space $(X, \tau)$ and a subset $F$ of $X$, is closed if the complement of $F$, $C(F)$ is open.
Some sets are neither open or closed. For example, in $\mathbb{R}$, the semi-open interval $[a, b)$ is neither open nor close.
Some sets are both open and closed. For example, in the subspace of $\mathbb{R}$ with $X = (x \lt 0) \cup (x \gt 0)$, we have that $(x \gt 0)$ is open and since its complement in $X$ is $(x \lt 0)$ is open, $(x \gt 0)$ is also closed!
A topological space $(X, \tau)$ is called a Hausforff space if for every pair of distinct elements $a, b \in X$, there exists neighborhoods of $a$ and $b$, $N$ and $M$ respectively, such that $N \cap M = \emptyset$.
Given a topological space $(X, \tau)$ and a subset $A$ of $X$, the closure of $A$, denoted by $\overline A$ is the set of elements $a$ such that for every neighborhood $N$ of $a$, $N \cap A \neq \emptyset$.
Intuitively the closure of a set $A$ is the set $A$ plus elements that are arbitrarily close to $A$. If $A$ is open these arbitrarily close points can be outside of itself. If $A$ is closed, then no, as the following Lemma states.
Lemma. $A$ is closed if and only if $A = \overline A$.
Another characterization of the closure of $A$ is the intersection of all closed sets that contain $A$, via the following Theorem:
Theorem. Let $Z$ be the set of all closed sets containing $A$. Then $\overline A = \bigcap_{F \in Z} F$
Given a topological space $(X, \tau)$ and a subset $A$ of $X$, the interior of $A$, denoted by $\mbox{Int}(A)$ is the set of elements $a$ for which $A$ is a neighborhood.
Intuitively the interior of $A$ is a subset of $A$ excluding elements that are at the “border” of $A$.
Another characterization of the interior of $A$ is the union of all open sets contained in $A$, via the following Theorem:
Theorem. Let $Z$ be the set of all closed sets containing $A$. Then $\mbox{Int}(A) = \bigcup_{O \in Z} O$
Given a topological space $(X, \tau)$ and a subset $A$ of $X$, the boundary of $A$, denoted by $\mbox{Bd}(A)$ is the set of elements in the closure of $A$ and the closure of its complement, that is,
\[\mbox{Bd}(A) = \overline{A} \cap \overline{C(A)}\]We might think that we can obtain $\mbox{Bd}(A)$ via $\overline{A} \setminus A$ but this only works if $A$ is open. If $A$ is closed however, $C(A)$ is open, so the boundary could be obtained via $\overline{C(A)} \setminus C(A)$. The definition using intersection accounts for these two cases.
A more intuitive definition is that of the difference between the closure and the interior:
\[\mbox{Bd}(A) = \overline{A} \setminus \mbox{Int}(A)\]Given a topological space $(X, \tau)$ and a subset $S$ of $X$, $x \in S$ is a isolated point if there exists a neighborhood of $x$ that contains no other points of $S$.
The definition is equivalent to the one for Metric Spaces (see Metric Spaces > Limit Point). Given a topological space $(X, \tau)$ and a subset $S$ of $X$, $x \in S$ is a limit point or accumulation point if every neighborhood of $x$ contains a point in $S$ other than $x$.
Given a topological space $(X, \tau)$ and a subset $S$ of $X$, $S$ is a dense set for every $x \in X$, it’s either in $S$ or arbitrarily close to a member of $A$.
Example. The set of rationals are dense in the reals, since for any $x \in \mathbb{R}$ either $x$ is rational or for any $\delta > 0$, there exists a rational $q$ such that $\abs{x - q} < \delta$.
A function maps one topological space $(X, \tau)$ into another $(Y, \tau’)$. If $a \in X$, then $f(a) \in Y$. If $A$ is a subset of $X$, then $f(A)$ is a subset of $Y$.
We can define the inverse function for a set $B \in Y$, denoted by $f^{-1}(B)$, as the subset $A’$ of $X$ such that $f(A’) = B$, that is:
\[f^{-1}(B) = \curly{a \in X : f(a) \in B}\]Note that if distincts $a, b \in X$ could have $f(a) = f(b) \in Y$. This leads to the following Lemma:
Lemma. $A \subseteq f^{-1}(f(A))$.
A function $f:X \rightarrow Y$ is surjective or onto if $Y = f(X)$, or that for every $y \in Y$ there is $x \in X$ such that $f(x) = y$. Note that the opposite, that for every $x \in X$ there is $f(x) \in Y$, is implicit in the definition of a function.
A function $f:X \rightarrow Y$ is injective or one-to-one if for every $f(a) = f(b) \in Y$, then $a = b \in X$. In other words, no two distinct elements in $X$ map to the same element in $Y$.
A function $f:X \rightarrow Y$ is bijective or one-to-one correspondence if it’s surjective and injective.
A function $f:(X, \tau) \rightarrow (Y, \tau’)$ is said to be continuous at point $a \in X$ if for every neighborhood of $N$ of $f(a)$, $f^{-1}(N)$ is a neighborhood of $a$. $f$ is continuous if it’s continuous for all points in $X$.
An equivalent definition: $f:(X, \tau) \rightarrow (Y, \tau’)$ is continuous if and only if for every $U$ that is an open set in $Y$, $f^{-1}(U)$ is an open set in $X$.
Topological spaces $(X, \tau)$ and $(Y, \tau’)$ are called homeomorphic if there exist inverse functions $f:X \rightarrow Y$ and $g:Y \rightarrow X$ and $f$ and $g$ are continuous.
Functions $f$ and $g$ are called homeomorphisms and they define a homeomorphism between $(X, \tau)$ and $(Y, \tau’)$.
Another characterization: $(X, \tau)$ and $(Y, \tau’)$ are homeomorphic if there exists a bijective function $f:X \rightarrow Y$ such that for every open set $O$ in $X$, $f(O)$ is open in $Y$.
Let $X$ and $Y$ be homeomorphic topological spaces. If $X$ having a property implies $Y$ having that property and vice-versa, such property is called a topological property.
Examples include connectedness and path-connectedness.
Let $\tau_1$ and $\tau_2$ be topologies on a given set $X$. $\tau_1$ is said to be weaker than $\tau_2$ if $\tau_1 \subset \tau_2$.
The product of two sets $A$ and $B$, denoted by $A \times B$ is the set of pairs corresponding to all combinations of elements from $A$ and $B$, that is $\forall a \in A, b \in B : (a, b) \in A \times B$.
The product of multiple sets $X_i$, $1 \le i le n$ can be denoted as $X = \prod_{i=1}^{n} X_i$. Every element $a \in X$ can be written as $a = (a_1, \cdots, a_n)$, where $a_i \in X_i$.
Let $(X_i, \tau_i)$, $1 \le i le n$ be topological spaces. Let $X = \prod_{i=1}^{n} X_i$. Let $O = \prod_{i=1}^{n} O_i$ where $O_i$ is some open set in $X_i$.
Let $\tau$ be the collection of subsets of $X$ that are unions of sets in the form of $O$ above. It’s possible to show $\tau$ is a topology, and the topological space $(X, \tau)$ is defined as the product of the topological spaces $(X_i, \tau_i)$, $1 \le i le n$.
Let $X_i$, $1 \le i \le n$ be sets and $X = \prod_{i=1}^{n} X_i$. Let $a = (a_1, \cdots, a_n) \in X$.
The function $p_i:X \rightarrow X_i$ called the $i$-th projection is defined as $p_i(a) = a_i$. If each $a_i$ are sets, then we can defined $p^{-1}_i(a_i) = X_1 \times \cdots \times a_i \times \cdots \times X_n$, that is, we’re “fixing” the $i$-th coordinate to be $a_i$ but leaving the others unrestricted.
Let $(X, \tau)$ and $(Y, \tau’)$ topological spaces. An identification is a continuous function $f:X \rightarrow Y$ if for each subset $U$ of $Y$, $f^{-1}(U) \in X$ being open in $X$ implies $U$ being open in $Y$.
Continuity only implies: if $U$ open set then $f^{-1}(U)$ open set. Identification adds the converse: if $f^{-1}(U)$ open set then $U$ open set.
Let $f: X \rightarrow Y$ be a surjective function. Define $\tau’$ as the set of subsets $U \in Y$ such that $f^{-1}(U)$ is open in $X$. It’s possible to show $\tau’$ is a topology of $Y$. This in turn proves that $f$ is continuous since if $U \in \tau’$ by construction it was added there because $f^{-1}(U)$ is an open set.
It’s also that case that for every $U \in Y$ if $f^{-1}(U)$ is open, then by construction $U \in \tau’$, so this function is an identification. We say that $\tau’$ is an identification topology on $Y$ determined by $f$.
A topological space is connected if it cannot be defined as the union of two disjoint non-empty open sets. If it can, then it’s disconnected.
An example of a disconnected space in $\mathbb{R}$: $x \lt 0 \cup x \gt 0$, since this set is the literal union of two disjoint non-empty open sets. On the other hand, the set $x \gt 0$ is connected. We can express it as the union of the disjoint sets $0 \lt x \le a$ and $x \ge a$ but the second set is not open.
Formally, let $(X, \tau)$ be a topological space. Then it is connected if no sets $A$ and $B$ exist such that:
If an open set $Y \in \tau$ also satisfies the above conditions, it’s called a connected open set. If $Y \neq \emptyset$ it’s also known as a region or domain.
A component of $a \in X$, denoted by $\mbox{Cmp}(a)$, is a maximal subset of $X$ that contains $a$ and is connected.
A topological space $X$ is locally connected at point $a$ if every neighborhood of $N$ of $a$ contains a neighborhood of $a$ that is connected.
A topological space $X$ is locally connected if it’s locally connected at each of its points.
Let $X$ be a topological space. A continuous function $f:[0, 1] \rightarrow X$ is called a path in $X$. $f(0)$ is called the initial point and $f(1)$ is called the terminal point. $f$ is said to connect points $f(0)$ and $f(1)$.
$X$ is path-connected if for every pair of points $a, b \in X$, there is a path that connects $a$ and $b$.
A simply connected space is a path-connected space where every path between two points can be continuously transformed into any other such path while preserving those points.
Intuitively a simply connected space is a connected space that has no holes, because if we consider two paths on different sides of the hole, they cannot be transformed into another continuously.
A loop is a path whose endpoints coincide. In a simply connected space, every loop can be contracted (transformed continuously) into a point. Again, in the presence of holes, a loop surrounding such hole cannot be turned into a point.
Let $X$ be a topological space, $Y$ a subset of $X$ and $\curly{A_\alpha}_{\alpha \in I}$ an indexed family of subsets of $X$ (See Set Theory). $\curly{A_\alpha}_{\alpha \in I}$ is called a covering of $Y$ if every element of $Y$ belongs to one of the $A_\alpha$, in other words: $Y \subseteq \cup_{\alpha \in I} A_\alpha$. If $I$ is finite, $\curly{A_\alpha}_{\alpha \in I}$ is called a finite covering of $Y$.
Let $\curly{A_\alpha}_{\alpha \in I}$ and $\curly{B_\beta}_{\beta \in J}$ be coverings of $Y$. If for every $A_\alpha$ there is $\beta \in J$ such that $A_\alpha = B_\beta$, then $\curly{A_\alpha}_{\alpha \in I}$ is a subcovering of $\curly{B_\beta}_{\beta \in J}$.
Note that we can’t define subcovering as $I \subseteq J$ because $A_\alpha$ and $B_\beta$ might be indexed differently (e.g. $A_\gamma \neq B_\gamma$ even if $\gamma \in I$, $\gamma \in J$).
Let $\curly{A_\alpha}_{\alpha \in I}$ be a covering of $Y$. If every $A_\alpha$ is an open subset of $X$ then $\curly{A_\alpha}_{\alpha \in I}$ is a open covering of $Y$.
Let $X$ be a topological space. It’s said to be compact if for every open covering $\curly{A_\alpha}_{\alpha \in I}$ of $X$, there exists a finite covering subcovering $\curly{A_\beta}_{\beta \in J}$ of $X$.