A metric space is a pair \((X,d)\) where \(X\) is a set and \(d\colon X\times X\to [0,\infty)\) is a function satisfying the following:
\(d(x,y)=0\) if and only if \(x=y\).
\(d(x,y)=d(y,x)\) for all all \(x,y\in X\).
\(d(x,y)\leq d(x,z)+d(z,y)\) for all \(x,y,z\in X\).
We call \(d\) a metric on \(X\).
Let \(p \in [1,\infty)\). \({\mathbb R}^n\) has a metric \[d_p(x,y)=({ \vert x_1-y_1 \vert^p+\cdots + \vert x_n-y_n \vert^p})^{1/p} = \left( \sum_{i=1}^n \vert x_i-y_i \vert^p\right)^{\frac{1}{p}}\]
\({\mathbb R}^n\) also has the metric \[d_\infty(x,y) = \max\{|x_i-y_i|\,|\,i=1,\ldots,n\}.\]
Let \(C([a,b])=\{f\colon [a,b]\to {\mathbb R}\,|\,f \mbox{ is continuous}\}\). Then \[d(f,g)=\sup \{|f(x)-g(x)|\,|\,x\in [a,b]\}\] is a metric on \(C([a,b])\).
Let \(X\) be any set. Then \[d(x,y)=\left\{ \begin{array}{cl} 0 & \mbox{if }x=y \\ 1 & \mbox{if }x\not=y \end{array} \right.\] is a metric, called the discrete metric on \(X\).
Let \((X,d)\) be a metric space, \(r>0\) and \(x\in X\). Then \[B(x;r)=\{y\in X\,|\,d(x,y)<r\}\] is called the open ball of radius \(r\) around \(x\in X\), and \[D(x;r)=\{y\in X\,|\,d(x,y)\leq r\}\] is called the closed ball of radius \(r\) around \(x\in X\).
One can draw the open and closed balls \(B(0;1)\) and \(D(0;1)\) for \({\mathbb R}^2\) for any of the metrics (1), (2), or (4) discussed above.
Let \(X\) be a metric space. A subset \(U\subset X\) is called open, if for every \(x\in U\) there is \(\varepsilon>0\) such that \(B(x;\varepsilon)\subset U\). A subset \(A\subset X\) is called closed, if \(X \setminus A\) is open.
Suppose \(X\) is a metric space and \(x \in N \subseteq X\). We call \(N\) an neighborhood of \(x\) if there exists an open set \(V \subseteq X\) with \(x \in V \subseteq N\).
Open balls are open, and closed balls are closed.
Let \((X,d)\) be a metric space.
A sequence in \(X\) is a function \(a\colon {\mathbb N}=\{0,1,2,3,\ldots\} \to X\). We write \((a_n)_{n\in {\mathbb N}}\) for the sequence, where \(a_n=a(n)\).
Let \(a\in X\). Then the sequence \((a_n)_{n\in{\mathbb N}}\) converges to \(a\), if for all \(\varepsilon>0\) there is \(n_0\in {\mathbb N}\) with \(d(a,a_n)<\varepsilon\) for all \(n\geq n_0\). We write \(\lim\limits_{n\to\infty}a_n=a\) or \(a_n\to a\).
Suppose that \(X\) and \(Y\) are metric spaces, \(a \in X\), and \(f: X \rightarrow Y\) is a function. Then the following are equivalent.
For all \(\epsilon > 0\) there exists \(\delta > 0\) such that \[d(a,x) < \delta \,\, \implies \,\, d(f(a), f(x)) < \epsilon {\rm ,}\]
For all \((a_n)\) with \(a_n \rightarrow a\), we have \(f(a_n) \rightarrow f(a)\).
For all open \(U \subseteq Y\) with \(f(a) \in U\), we have that \(f^{-1}(U)\) is a neighborhood of \(a\).
For a function \(f\) as in the previous proposition satisfying the first of the (equivalent) conditions, we say that \(f\) is continuous at \(a\).
If \(f: X \rightarrow Y\) is continuous at all points \(a \in X\) we say that \(f\) is continuous.
A function \(f : X \rightarrow Y\) is continuous if and only if \(f^{-1}(U)\) is open for all open \(U \subseteq Y\).
Any continuous function \(f\colon {\mathbb R}\to {\mathbb R}\) that you have seen in Calculus.
1 Define \(I\colon C[0,1] \to C[0,1]\) by \[I(f) (x) = \int\limits_0^x f(t) dt\] This is continuous with the metric from Example 1.2 (3).
Let \(X\) be a set with the discrete metric, and let \(f\colon X \to {\mathbb R}\) be a function. Then \(f\) is continuous. On the other hand, \(\mathop{\mathrm{id}}\colon ({\mathbb R},d_2)\to ({\mathbb R},d)\) with \(d\) the discrete metric on \({\mathbb R}\) is not continuous.
Let \(X\) be a metric space.
\(X\) and \(\emptyset\) are both open and closed.
an arbitrary union of open sets is open.
a finite intersection of open sets is open.
a finite union of closed sets is closed.
an arbitrary intersection of closed sets is closed.
Let \(M={\mathbb R}^2\) with the euclidean metric. Then \(\{0\}\) is closed, but not open. Also, \[\{0\}=\bigcap_{i=1}^\infty B(0;\frac{1}{i})\] so the arbitrary intersection of open sets need not be open.
\(D(0;1)\) is closed, but not open.
If \((X,d)\) is discrete, then any \(A\subset X\) is open and closed.
For any set \(X\) we write \(\wp(X) = \{ A : A \subseteq X \}\) for the power set of \(X\), that is, for the set whose elements are all the subsets of \(X\).
Let \(X\) be a set. A topology on \(X\) is a subset \(\tau \subseteq \wp(X)\) which satisfies the following:
\(\emptyset\) and \(X\) are in \(\tau\).
If \(U_i\in \tau\) for all \(i\in I\), then \(\bigcup_{i\in I}U_i\in \tau\).
If \(U_1, U_2\in \tau\), then \(U_1\cap U_2\in \tau\).
The pair \((X,\tau)\) is called a topological space. The elements of \(\tau\) are called open subsets of \(X\). A subset \(A\subset X\) is called closed, if its complement \(X \setminus A\) is open.
Let \((M,d)\) be a metric space, and \(\tau_d\) the collection of open sets in the sense of Definition [def:open]. Then \((M,\tau_d)\) is a topological space.
Let \(X\) be any set, and \(\tau=\wp(X)\). This is called the discrete topology on \(X\).2
Let \(X\) be any set, and \(\tau=\{\emptyset, X\}\). This is called the indiscrete topology on \(X\).3
Let \(X=\{0,1,2,3,\ldots\}\) and \(\tau=\{\emptyset\}\cup \{U\subseteq X\,|\, (X \setminus U) \mbox{ is finite}\}\).
Let \((X,\tau)\) be a topological space, \(A\subset X\). Then \[\tau_A=\{A\cap U\,|\, U\in \tau\}\] is a topology, called the induced, or subspace topology. Note that \({\mathbb Z}\subset {\mathbb R}\) is discrete in the subspace topology, and \({\mathbb Q}\subset {\mathbb R}\) is not discrete in the subspace topology.
On \({\mathbb R}^n\) we have the metrics \(d_p\) for \(p \in [1, \infty)\), and also \(d_\infty\). They all induce the same topology on \({\mathbb R}^n\). This is called the standard topology on \({\mathbb R}^n\).
Let \(X\) be a topological space. Then \(X\) and \(\emptyset\) are closed. Furthermore, arbitrary intersections of closed sets are closed, and finite unions of closed sets are closed.
A topological space \(X\) is called Hausdorff, if whenever \(x,y\in X\) with \(x\not=y\), there exist open sets \(U,V\subseteq X\) with \(x\in U\), \(y\in V\) and \(U\cap V=\emptyset\).
Let \(M\) be a metric space. Then \(M\) is a Hausdorff space.
Let \(X\) be a set with more than one element, and give it the indiscrete topology. Then \(X\) is not a Hausdorff space.
We give Furstenberg’s topology on \({\mathbb Z}\). A subset \(U \subseteq {\mathbb Z}\) is defined to be open if for every \(a \in U\) there exists a non-zero \(d \in {\mathbb Z}\) with \(a + d{\mathbb Z}\subseteq U\).
This is also a Hausdorff topology. Is it induced from a metric on \({\mathbb Z}\)?
Let \(X,Y\) be topological spaces. A function \(f\colon X\to Y\) is called continuous, if for every open subset \(U\subset Y\) the inverse image \(f^{-1}(U)\) is open in \(X\). Let \(x\in X\). The function \(f\) is called continuous at \(x\), if for every open set \(U\subseteq Y\) with \(f(x)\in U\) there is an open set \(V\subseteq X\) with \(x\in V\) such that \(f(V)\subseteq U\).
A function \(f : X \rightarrow Y\) is continuous iff it is continuous at all points of \(X\).
Let \(X\) be a topological space and \(A\subseteq X\). The inclusion \(i\colon A \hookrightarrow X\) is continuous, if \(A\) is given the subspace topology.
This generalizes all previously met notions of continuity. Any previously met notion is (really) about continuity of maps between metric spaces, and this is equivalent to continuity of maps with respect to the induced topologies.
Composition of continuous functions is a continuous function.
Let \(f\colon X \to Y\) be a function between topological spaces \(X\) and \(Y\). Then \(f\) is continuous if and only if \(f^{-1}(A)\) is closed for every \(A\subseteq Y\) closed.
The \(n\)-sphere is defined as \[S^n=\{(x_1,\ldots,x_{n+1})\in {\mathbb R}^{n+1}\,|\,x_1^2+\cdots + x_{n+1}^2=1\}\] and is a closed subset of \({\mathbb R}^{n+1}\).
Denote by \(M_{n,n}({\mathbb R})\) the set of \(n\times n\)-matrices, topologized as \({\mathbb R}^{n^2}\). Then \(\mathrm{GL}_n({\mathbb R})\), the set of invertible \(n\times n\)-matrices is an open subset of \(M_{n,n}({\mathbb R})\). Also, \(\mathrm{SL}_n({\mathbb R})=\{A\in \mathrm{GL}_n({\mathbb R})\,|\, \det A=1\}\) is a closed subset of \(M_{n,n}({\mathbb R})\).
The orthogonal group is defined as \[\mathrm{O}(n)=\{A\in \mathrm{GL}_n({\mathbb R})\,|\, AA^t=I\}\] which is also a closed subset. The special orthogonal group is \[\mathrm{SO}(n)= \mathrm{O}(n)\cap \mathrm{SL}_n({\mathbb R})\] and it is closed as an intersection of closed sets.
Let \(X,Y\) be topological spaces. A map \(h\colon X\to Y\) is called a homeomorphism, if \(h\) is bijective and \(h^{-1}\) is continuous. In that case \(X\) and \(Y\) are called homeomorphic, and we write \(X\approx Y\). Note that \(h\) induces a bijection between \(\tau_X\) and \(\tau_Y\) (and this bijection commutes with unions and intersections).
The interval \((0,1)\) is homeomorphic to \({\mathbb R}\). We will see later that \([0,1)\) is not homeomorphic to \({\mathbb R}\).
\({\mathbb R}\) with the standard topology is not homeomorphic to \({\mathbb R}\) with the discrete topology.
We give an example of a non-Hausdorff topological space \(X\) in which every point \(a \in X\) sits inside an open set \(a \in U \subseteq X\) where \(U\) is homeomorphic to \({\mathbb R}\) in the standard topology. This space is known as the real line with the origin doubled.
To close this section we return to the earlier-defined Furstenberg topology on \({\mathbb Z}\).
We consider the subset \[X := \bigcup_{p \in {\mathbb Z}, \, p \,\, {\rm prime}} p {\mathbb Z}\] and use this to show that there are infinitely many prime numbers.
Let \(X\) be a topological space and \(x\in X\), \(A\subseteq X\).
An open neighborhood of \(x\) is an open set \(N\) containing \(x\).
A point \(x\in X\) is called a limit point of \(A\), if every open neighborhood \(N\) of \(x\) satisfies \((N \setminus \{x\})\cap A \not= \emptyset\).
\(X={\mathbb R}\), \(0\in X\), \((-\frac{1}{2},\frac{1}{2})\) is an open neighborhood of \(0\).
If \(U\subseteq X\) is open, then \(U\) is an open neighborhood for every \(x\in U\).
Let \(A=\{\frac{1}{n}\in {\mathbb R}\,|\, n\in {\mathbb Z}\setminus \{0\}\}\). Then \(A\) has exactly one limit point \(0\), and \(0\notin A\).
Let \(X\) be a topological space, \(A\subseteq X\). The interior of \(A\), denoted \(A^\circ\), is the largest open set contained in \(A\). The closure of \(A\), denoted \(\bar{A}\), is the smallest closed set which contains \(A\). More precisely, \[A^\circ=\bigcup_{U\subseteq A \mbox{ open}} U \hspace{1cm} \bar{A}=\bigcap_{C\supset A \mbox{ closed}} C\] A subset \(A\subseteq X\) is called dense, if \(\bar{A}=X\).
Let \({\mathbb Q}\subseteq {\mathbb R}\), then \({\mathbb Q}^\circ=\emptyset\) and \(\bar{{\mathbb Q}}={\mathbb R}\).
Let \(A\subseteq X\). Then \(\bar{A}=A\cup \{ \mbox{limit points of }A \}\).
Let’s consider again Furstenberg’s topology on \({\mathbb Z}\). Let \(A = \{ p \in {\mathbb Z}: p \,\, {\rm prime} \} \subseteq {\mathbb Z}\). What is \(\bar{A}\)? (We might need to borrow Dirichlet’s Theorem from number theory to answer this).
Let \(M\) be a metric space, \(A\subseteq M\) and \(x\) a limit point of \(A\). Then there exists a sequence \(x_n\in A\) such that \(\lim x_n=x\). Furthermore, if \(x\in M\setminus A\) and there is a sequence \(x_n\in A\) with \(\lim x_n = x\), then \(x\) is a limit point of \(A\).
Let \(X\) be a topological space. A basis \(\mathcal{B}\) for the topology of \(X\) is a collection of open sets such that every open set \(U\) is the union of elements of \(\mathcal{B}\).
Let \(M\) be a metric space. The collection \(\{B(x;r)\,|\,x\in M, R>0\}\) is a basis for the topology of \(M\).
Let \(X={\mathbb R}^n\) with the standard topology. The collection
\(\{B(q;\frac{1}{m})\,|\,q\in {\mathbb Q}^n, m\geq 1 \mbox{ integer}\}\) is a countable basis of the topology.
Let \(f\colon X\to Y\) be a function between topological spaces. The following are equivalent.
\(f\) is continuous.
If \(\mathcal{B}\) is a basis for the topology of \(Y\), then \(f^{-1}(B)\) is open for every \(B\in \mathcal{B}\).
\(f(\bar{A})\subseteq \overline{f(A)}\) for every subset \(A\subseteq X\).
\(\overline{f^{-1}(B)} \subseteq f^{-1}(\bar{B})\) for every subset \(B\subseteq Y\).
\(f^{-1}(B)\) is closed for every closed set \(B\subseteq Y\).
A basis for a topology usually gives a collection of open sets which are somewhat easy to handle. By [theorem:basis](2) we see that continuity is captured by a basis. Another situation is that one may have a set \(X\) and a collection \(\mathcal{B}\) of subsets that one would like to be open – one would like a topology for which \(\mathcal{B}\) is a basis.
Let \(X\) be a set and \(\mathcal{B}\) a collection of subsets of \(X\) such that
For each \(x\in X\) there is a \(B\in \mathcal{B}\) with \(x\in B\).
If \(x\in B_1\cap B_2\), where \(B_1,B_2\in \mathcal{B}\), then there is a \(B_3\in \mathcal{B}\) with \(x\in B_3\subseteq B_1\cap B_2\).
Then there is a unique topology \(\tau_\mathcal{B}\) on \(X\) of which \(\mathcal{B}\) is a basis.
Let \(X\), \(Y\) be topological spaces. The cartesian product of \(X\) and \(Y\) is the set \[X\times Y = \{(x,y)\,|\,x\in X, y\in Y\}\] and it is given the product topology, obtained through Theorem [theorem:gener] by using \[\mathcal{B}_{X\times Y}=\{U\times V\,|\, U\subseteq X, V\subseteq Y, \, U \,\, {\rm open}, \, V \,\, {\rm open} \}.\]
Let \(X={\mathbb R}\), \(Y={\mathbb R}\). Then \(X\times Y= {\mathbb R}^2\), and the product topology agrees with the standard topology.
Let \(X=S^1=Y\). Then \(T^2=S^1\times S^1\) is called the torus. We can also form the \(n\)-torus by defining \[T^n= S^1\times T^{n-1},\] inductively.
If \(X\) is a set and \(\tau\) and \(\tau'\) are topologies on \(X\), we say that \(\tau\) is smaller than \(\tau'\) (or that \(\tau'\) is larger than \(\tau\)) iff \(\tau \subseteq \tau'\).
Let \(X\), \(Y\) be topological spaces, and \(X\times Y\) given the product topology. The projections \(p_X\colon X\times Y \to X\) and \(p_Y\colon X\times Y \to Y\) are both continuous and map open sets to open sets. Furthermore, the product topology is the smallest topology so that both projections are continuous.
Let \(X,Y,Z\) be topological spaces. A function \(f\colon Z \to X\times Y\) is continuous if and only if both \(p_X\circ f\colon Z\to X\) and \(p_Y\circ f\colon Z\to Y\) are continuous.
Let \(f\colon X\to {\mathbb R}^n\) be a function. We then have \(n\) coordinate functions \(f_1,\ldots,f_n\colon X\to {\mathbb R}\), and \(f\) is continuous if and only if these coordinate functions are continuous.
Let \(X\), \(Y\) be non-empty topological spaces. Then \(X\times Y\) is a Hausdorff space if and only if \(X\) and \(Y\) are Hausdorff spaces.
If \(X\) is a topological space then we call a subset \(A \subset X\) clopen iff it is both closed and open.
Note that in any topological space \(X\), the subsets \(\phi\) and \(X\) are both clopen.
A topological space is called connected iff the only clopen subsets of \(X\) are \(\phi\) and \(X\).
\({\mathbb R}\) is connected.
\({\mathbb Q}\) is not connected.
The connected subsets of \({\mathbb R}\) are the intervals.
Let \(X=\{0,1\}\) with the discrete topology. Then \(X\) is not connected.
If the topology of \(X\) is given by \(\tau = \{\emptyset, \{1\},\{0,1\}\}\), then \(X\) is connected.
The Furstenberg topology on \({\mathbb Z}\) is not connected.
The continuous image of a connected space is connected. In particular, if \(h\colon X\to Y\) is a homeomorphism with \(X\) connected, then \(Y\) is connected.
Let \(X\) be a topological space. The following are equivalent.
\(X\) is connected.
\(X\) cannot be written as the union of two disjoint, non-empty open sets.
There is no continuous surjective function from \(X\) to a discrete space with more than one point.
\(\det\colon GL_n({\mathbb R})\to {\mathbb R}-\{0\}\) is onto. Hence \(GL_n({\mathbb R})\) is not connected.
\(O(n)\) is not connected.
\((0,1)\) is connected.
\(X=(0,1)\) and \(Y=(0,1]\) are not homeomorphic.
Let \(X\) be a topological space, \(Z\subseteq X\) connected and \(Y\subseteq X\) with \(Z\subseteq Y\subseteq \bar{Z}\). Then \(Y\) is connected. In particular, the closure of a connected set is connected.
Let \(X\) be a topological space and \(Z\subseteq X\). If \(Z\) is connected and \(\bar{Z}=X\), then \(X\) is connected.
Let \(\mathcal{A}=\{A_i\,|\,i\in I\}\) be a collection of subsets of \(X\) such that \(\bigcup_{i\in I}A_i=X\). Assume that each \(A_i\) is connected and for each \(i,j\in I\) we have \({A}_i\cap {A}_j\not=\emptyset\). Then \(X\) is connected.
If \(X\) and \(Y\) are connected, then \(X\times Y\) is connected.
\({\mathbb R}^n\) is connected.
\(B^n\) is connected.
\(D^n\) is connected.
\(S^n\) is connected for \(n\geq 1\).
\(T^n\) is connected for \(n\geq 1\).
A component of a topological space \(X\) is a maximal connected subset of \(X\).
Each component of a topological space is a closed subset and if \(C_1\) and \(C_2\) are different components, then \(C_1\cap C_2=\emptyset\). Also, a topological space is the union of its components.
If \(X\) is connected, there is only one component.
If \(X\) is discrete, each point is a component.
In \({\mathbb Q}\) every point is a component.
A path in a topological space \(X\) is a continuous function \(\gamma\colon [0,1]\to X\), we say \(\gamma\) is a path from \(\gamma(0)\) to \(\gamma(1)\).
A topological space is called path connected, if for any two points \(x,y\in X\) there is a path from \(x\) to \(y\).
A path connected space is connected.
Let \[Z = \{(x,\sin \frac{\pi}{x})\in {\mathbb R}^2\,|\,0<x\leq 1\}.\] Then the closure \(\overline{Z}\) is connected, but not path connected.
Let \(X\) be a topological space. A cover of \(X\) is a collection \((U_i)_{i\in I}\) of subsets of \(X\) with \(\bigcup_{i\in I}U_i=X\). The cover is called an open cover, if every \(U_i\) is open. A subcover of \((U_i)_{i\in I}\) is a cover \((U_j)_{j\in J}\) where \(J\subset I\).
The topological space \(X\) is said to be compact if and only if every open cover of \(X\) admits a finite subcover.
If \(X\) is a finite set, then \(X\) is compact.
\({\mathbb R}\) is not compact.
Let \(X\) be an infinite set with the topology \(\tau=\{U\subset X\,|\,X \setminus U\mbox{ is finite}\}\cup\{\emptyset\}\). Then \(X\) is compact.
The closed interval \([0,1]\) is compact.
The continuous image of a compact space is compact.
A closed subset of a compact space is compact.
A compact subset of a Hausdorff space \(X\) is closed in \(X\).
A bijective continuous function from a compact space to a Hausdorff space is a homeomorphism.
The integers \({\mathbb Z}\) with the Furstenberg topology is not compact.
Let \(X,Y\) be compact spaces. Then \(X\times Y\) is compact.
[Heine-Borel] A subset of \({\mathbb R}^n\) is compact if and only if it is closed and bounded.
\(S^n\) is compact.
The torus \(T^n=S^1\times\cdots\times S^1\) is compact.
\(X=\{x\in {\mathbb R}^3\,|\,x_1^2+x_2^2-x_3^3=1\}\) is not compact.
In a metric space the equivalence between compact sets and closed and bounded sets usually does not work.
Let \(f\colon X\to {\mathbb R}\) be a map from a compact space to the reals. Then \(f\) attains its maximum and minimum.
[Bolzano-Weierstrass] An infinite subset of a compact space must have a limit point.
Let \(X\) be a set and \(\sim\) an equivalence relation on \(X\). We define \(X/\sim\) to the set of equivalence class of \(\sim\). For \(x \in X\) we have \[[x] = \{ y \in X : y \sim x \}\] and \[X/ \sim = \{ [x] : x \in X \} { \rm . }\]
Note that there is a surjective map \[\pi : X \rightarrow X / \sim : x \mapsto [x]\] which takes elements of \(X\) to their equivalence class.
Suppose \(X\) is a topological space and \(\sim\) is an equivalence relation on \(X\). Then we give \(X / \sim\) the topology defined by
\(U \subseteq X/\sim\) is open if and only if \(\pi^{-1} (U)\) is open in \(X\).
This defines a topology on \(X / \sim\).
The topology on \(X / \sim\) is the largest such that \(\pi\) is continuous.
If \(A \subseteq X\) then define the equivalence relation \(\sim\) by \(x \sim y\) if and only if either \(x=y\) or \(x,y \in A\). Then we define \(X/A\) to be \(X/\sim\).
Here’s a lemma that often use implicitly in proofs from now on.
Suppose \(A \subseteq Y\) with the subspace topology and let \(i : A \hookrightarrow Y\) be inclusion. Then \[f: X \rightarrow A\] is continuous if and only if \[i \circ f : X \rightarrow Y\] is continuous.
Consider \({\mathbb R}\) with its standard topology and the equivalence relation given by \(x \sim y\) iff \(x-y \in {\mathbb Z}\). Then (we shall see that) \({\mathbb R}/\sim\) is homeomorphic to the circle \(S^1\).
Consider the closed \(n\)-ball \(D^n \subseteq {\mathbb R}^n\). The \(n-1\)-sphere is a subset of this \(S^{n-1} \subseteq D^n\). We shall see that \(D^n/S^{n-1}\) is homeomorphic to \(S^n\).
Suppose \(X\) is a space with an equivalence relation \(\sim\), and \(Y\) is another space. Then a map \[f: X/\sim \rightarrow Y\] is continuous if and only if \[f \circ \pi : X \rightarrow Y\] is continuous.
A topological group \(G\) is a Hausdorff space \(G\) which is also a group, such that multiplication \(\cdot\colon G\times G \to G\) and inversion \(i\colon G \to G\) are continuous.
\({\mathbb R}^n\) with addition.
\({\mathbb C}\setminus \{0\}\) with multiplication. Then \(S^1\subseteq {\mathbb C}\) is a subgroup.
The set of invertible \(n\times n\) matrices \(\mathrm{GL}_n({\mathbb R})\), with subgroups \(\mathrm{O}(n)\) and \(\mathrm{SO}(n)\).
Any subgroup \(H\) of a topological group \(G\) is a topological group.
Let \(\mathbb{H}= {\mathbb R}^4 = \langle 1,i,j,k \rangle\), the set of quaternions, with topology from \({\mathbb R}^4\). We can define a multiplication by \[ij=k, ji=-k, jk = i, kj = -i, ki = j, ik = -j\] and extending by the usual distributive laws. This turns \(\mathbb{H}\setminus \{0\}\) into a (non-abelian) multiplicative group, and \(S^3 \subseteq {\mathbb R}^4 \setminus \{ 0 \}\) forms a subgroup.
Any group \(G\) with the discrete topology is a topological group.
Let \(G\) be a topological group. For \(x\in G\) define \(L_x\colon G\to G\) by \(L_x(g)=xg\), called left translation by \(x\). This is continuous, and has an inverse, namely \(L_{x^{-1}}\). In particular left translation is a homeomorphism. Similarly, right translation \(R_x\) is a homeomorphism.
Let \(G\) be a topological group and \(K\) the connected component of \(G\) which contains the identity element. Then \(K\) is a closed normal subgroup of \(G\).
We know that \(\mathrm{O}(n)\) is not connected, but we will see that \(\mathrm{SO}(n)\) is connected. In particular, \(\mathrm{SO}(n)\) is the \(K\) in the case of \(G=\mathrm{O}(n)\).
An action of a group \(G\) on a topological space \(X\) is a map \(\bullet : G \times X \rightarrow X\) such that
\((hg)\bullet x = h\bullet (g\bullet x)\) for all \(h,g\in G\) and \(x\in X\)
\(1\bullet x = x\) for all \(x\in X\).
\(g : X \rightarrow X\) defined by \(g(x) = g \bullet x\) is continuous.
Note that for \(g\in G\) the map \(x\mapsto g\bullet x\) is a homeomorphism with inverse coming from \(g^{-1}\).
In the case that \(G\) is a topological group we make a further requirement on the action:
An action of a topological group \(G\) on a topological space \(X\) is a continuous map \(\bullet\colon G\times X \to X\) such that
\((hg)\bullet x = h\bullet (g\bullet x)\) for all \(h,g\in G\) and \(x\in X\).
\(1\bullet x = x\) for all \(x\in X\).
Note that if one forgets the topological structure of \(G\), this is just a group action of \(G\) on \(X\).
Trivial action : \(g \cdot x = x\) for all \(g\in G\) and \(x\in X\).
\(G= \mathrm{GL}_n({\mathbb R})\), \(X={\mathbb R}^n\), the action given by matrix multiplication \(A\cdot x = Ax\). This induces an action of \(\mathrm{O}(n)\) and \(\mathrm{SO}(n)\) acting on \(S^{n-1}\).
Let \(G\) be a topological group and \(H\) a subgroup. Then \(H\) acts on \(G\) via \(H\times G \to G\) given by \(h\bullet g = hg\).
If \(N\) is a normal subgroup of \(G\), then \(G\) acts on \(N\) by \(g\bullet n = gng^{-1}\).
\(S^3\) acts on \({\mathbb R}^4\) by quaternion multiplication. This induces an injection \(T\colon S^3 \to \mathrm{SO}(4)\).
If \(G\) acts on \(X\), we can define an equivalence relation \(\sim\) on \(X\) by saying \(x\sim y\) if there is \(g\in G\) with \(gx=y\). An equivalence class is called an orbit, and denoted by \(Gx\). The corresponding quotient space is called the orbit space, denoted by \(X/G\). If \(X/G\) is just a point, the action is called transitive.
If \(G\) acts trivially, every orbit is a point and \(X/G = X\).
The action of \(\mathrm{O}(n)\) on \(S^{n-1}\) is transitive. The action of \(\mathrm{SO}(n)\) on \(S^{n-1}\) is transitive, provided that \(n\geq 2\).
\({\mathbb R}^n/\mathrm{GL}_n({\mathbb R})\) consists of two points, the orbit of \(0\), and a non-zero orbit. The topology on the quotient is neither discrete nor indiscrete.
Let \(G\) be a connected topological group that acts on a topological space \(X\) such that \(X/G\) is connected. Then \(X\) is connected.
If we include \(i: \mathrm{SO}(n) \hookrightarrow \mathrm{SO}(n+1)\) by \[i : A \mapsto \left( \begin{array}{cc} 1 & 0 \\ 0 & A \end{array} \right)\] then \(\mathrm{SO}(n)\) acts on \(\mathrm{SO}(n+1)\) as a subgroup. We have \[\mathrm{SO}(n+1) / \mathrm{SO}(n) = S^n {\rm .}\]
\(\mathrm{SO}(n)\) is connected for \(n\geq 1\).