Topological Spaces & Continuity
Topological Spaces & Continuity
Topology studies properties preserved under continuous deformations — stretching, bending — but not cutting or gluing. A topological space $(X,\mathcal{T})$ specifies which subsets are 'open', generalizing the open intervals of $\mathbb{R}$. Continuity, compactness, and connectedness are the central topological concepts. Two spaces are homeomorphic if there is a bicontinuous bijection between them — a topological isomorphism.
Topological Space & Continuous Maps
A topology $\mathcal{T}$ on $X$ is a collection of subsets (open sets) satisfying: $\emptyset,X\in\mathcal{T}$; arbitrary unions of open sets are open; finite intersections are open. Examples: discrete ($\mathcal{T}=2^X$), indiscrete ($\mathcal{T}=\{\emptyset,X\}$), metric topology (open balls generate $\mathcal{T}$), subspace topology. A map $f:(X,\mathcal{T})\to(Y,\mathcal{S})$ is continuous iff $f^{-1}(V)\in\mathcal{T}$ for all $V\in\mathcal{S}$. A homeomorphism is a bijective continuous map with continuous inverse.
Basis for a Topology
A basis $\mathcal{B}$ for a topology: every open set is a union of basis elements. Condition: (i) every $x\in X$ belongs to some $B\in\mathcal{B}$; (ii) if $x\in B_1\cap B_2$, $\exists B_3\ni x$ with $B_3\subseteq B_1\cap B_2$. The metric topology has basis $\{B(x,r):x\in X,r>0\}$. Product topology: for $X\times Y$, basis elements $U\times V$ with $U$ open in $X$, $V$ open in $Y$. Subbase: arbitrary intersections of subbase elements form a base.
Example 1
Show the standard topology on $\mathbb{R}$ equals the order topology.
Solution: Standard: generated by open intervals $(a,b)$. Order topology: generated by rays $(-\infty,b)$ and $(a,\infty)$. Each $(a,b)=(-\infty,b)\cap(a,\infty)$ is open in order topology; each ray is open in standard topology. So both topologies have the same open sets.
Example 2
Show $\mathbb{R}$ and $(0,1)$ are homeomorphic.
Solution: Map $f:(0,1)\to\mathbb{R}$: $f(x)=\tan(\pi(x-1/2))$ is a bijection, continuous, with continuous inverse $f^{-1}(y)=\frac{1}{\pi}\arctan(y)+\frac{1}{2}$. So $(0,1)\cong\mathbb{R}$ as topological spaces. Intuitively: the bounded interval can be 'stretched' to the entire line.
Practice
- Show that the union of two topologies on $X$ need not be a topology.
- Prove that a map $f:X\to Y$ is continuous iff the preimage of every closed set is closed.
- Describe the quotient topology and show $[0,1]/\{0,1\}\cong S^1$.
- Show that any map from a discrete space is continuous.
Show Answer Key
1. On $\{a,b,c\}$: $\tau_1=\{\emptyset,\{a\},X\}$ and $\tau_2=\{\emptyset,\{b\},X\}$ are topologies. But $\tau_1\cup\tau_2=\{\emptyset,\{a\},\{b\},X\}$ is not a topology since $\{a\}\cup\{b\}=\{a,b\}\notin\tau_1\cup\tau_2$.
2. ($\Rightarrow$) $f$ continuous: $f^{-1}(U)$ open for open $U$. If $C$ is closed, $C^c$ is open, so $f^{-1}(C^c)=(f^{-1}(C))^c$ is open, hence $f^{-1}(C)$ is closed. ($\Leftarrow$) Reverse the argument.
3. The quotient topology on $[0,1]/\{0,1\}$ identifies endpoints. Map $f:[0,1]\to S^1$ by $f(t)=e^{2\pi it}$. This is continuous, surjective, and respects the identification ($f(0)=f(1)$). It descends to a continuous bijection on the quotient. Since $[0,1]$ is compact and $S^1$ is Hausdorff, $f$ is a homeomorphism.
4. In a discrete space, every subset is open. So for any $f:X\to Y$ and open $U\subseteq Y$, $f^{-1}(U)\subseteq X$ is open (trivially). Hence $f$ is continuous.