A path in (or to) a topological space \(X\) is a continuous function \(f\colon [a,b]\rightarrow X\text{,}\) where \([a,b]\subseteq \R\) is a closed interval. We call \(f(a)=x\) the start point (or initial point) and \(f(b)=x'\) the end point (or terminal point) of the path \(f\text{.}\)
Definition1.13.2.Path-connected space.
Let \(X\) be a topological space. Two elements \(x, x'\in X\) are connected by a path in \(X\) if there is a path \(f\colon [a,b]\rightarrow X\) such that \(f(a)=x\) and \(f(b)=x'\text{.}\) The space \(X\) is path connected if any two elements of \(X\) are connected by a path.
Theorem1.13.3.Path-connected implies connected.
If \(X\) is path connected, then \(X\) is connected. Using logical shorthand:
\begin{equation*}
X \text{ path connected}\implies X \text{ connected}.
\end{equation*}
Proof.
Assume by contradiction that \(X=U_1\cup U_2\) is a separation of \(X\text{.}\) We can find points \(x_1\in U_1, x_2\in U_2\text{.}\) Let \(f\colon [a,b]\rightarrow X\) be a path connecting \(x, x'\text{.}\) By Theorem 1.12.7, the image \(f([a.b])\) is connected. By Theorem 1.12.6 we must have \(f([a,b])\subseteq U_1\) or \(f([a,b])\subseteq U_2\text{.}\) This contradicts the fact that \(f(a)=x_1\in U_1, f(b)=x_2\in U_2\) and \(U_1\cap U_2=\emptyset\text{.}\) Thus there can be no separation of \(X\text{,}\) and we conclude that it is connected.
Theorem1.13.4.Path-connectedness and continuity.
If \(f\colon X\rightarrow Y\) is continuous and \(X\) is path connected, then \(f(X)\) is path connected.
Proof.
Given \(y=f(x)\) and \(y'=f(x')\text{,}\) let \(\phi\colon [a,b]\rightarrow X\) be a path connecting \(x\) and \(x'\text{.}\) The composition \(f\circ \phi\) is then a path connecting \(y\) and \(y'\text{.}\)
Example1.13.5.Balls in \(\R^n\).
Let \(X=\R^n\) be Euclidean \(n\)-space. For all \(\epsilon >0\) and any \(\boldx\in \R^n\text{,}\) the open ball \(B_\epsilon(\boldx)\) is connected, as is the closed ball \(\overline{B}_\epsilon(\boldx)\) defined as
It is easy to see that the straight line path \(\phi\colon[0,1]\rightarrow \R^n\) defined as \(\phi(t)=(1-t)\boldx+t\boldy\) is a continuous function connecting any two points in \(\R^n\text{.}\) For our current purposes, we wish to show that if \(\boldy\in B_{\epsilon}(\boldx)\text{,}\) then so is \((1-t)\boldx+t\boldy\text{,}\) for any \(t\in [0,1]\text{.}\) To this end, note that
This proves that the open ball is path connected, and hence connected. Next, it is easy to see that the closed ball is in fact the closure of the open ball. Since connectedness is preserved under closure, we conclude that the closed ball is connected.
Example1.13.6.Punctured Euclidean space \(\R^n-\{\boldzero\}\text{:}\)\(n\geq 2\).
Let \(X=\R^n\) be Euclidean \(n\)-space, where \(n\geq 2\text{.}\) The subspace \(\R^n-\{\boldzero\}\) is connected.
Solution.
Take any \(\boldx,\boldy\in \R^n-\{\boldzero\}\text{.}\) If the two vectors are not scalar multiples of one another, then the straight line path connecting them is guaranteed to lie in \(\R^n-\{\boldzero\}\text{,}\) using a linear independence argument. If \(\boldy=c\boldx\text{,}\) then since \(\dim \R^n\geq 2\text{,}\) we can find a vector \(\boldz\in \R^n-\{\boldzero\}\) that is not a scalar multiple of either vector. We can then construct a polygonal path lieing in \(\R^n-\{\boldzero\}\) that first connects \(\boldx\) to \(\boldz\) via a straight line path, and then connects \(\boldz\) to \(\boldy\) via a straight line path.
Example1.13.7.\(\R\not\cong \R^n\) for \(n\geq 2\).
Prove: if \(n\geq 2\text{,}\) then there is no homeomorphism from \(\R\) to \(\R^n\text{.}\)
Solution.
Assume by contradiction that \(\phi\colon \R\rightarrow \R^n\) is a homeomorphism for some \(n\geq 2\text{.}\) It follows that the restriction \(\phi\colon \R-\{0\}\rightarrow \R^n-\{\phi(0)\}\) is a homeomorphism. This a contradiction since \(\R-\{0\}\) is not connected (\((0,\infty)\) is open and closed in \(\R-\{0\}\)), but \(\R^n-\{\phi(0)\}\) is connected.
Example1.13.8.The \(n\)-sphere.
Fix \(n\geq 1\text{.}\) The \(n\)-sphere \(S^n\) is defined as
Prove: \(S^n\) is connected for all \(n\geq 1\text{.}\)
Solution.
Fix \(n\geq 1\) and consider the map \(\phi\colon \R^{n+1}-\{\boldzero\}\rightarrow S^n\) defined as \(\phi(\boldx)=\frac{\boldx}{\norm{\boldx}}\text{,}\) where as usual
This map is continuous (homework exercise!) and surjective. Since \(\R^{n+1}-\{\boldzero\}\) is connected, we see that \(S^n\) is the image of a connected set under a continuous function, hence connected.
Topological specimen15.Topologist’s sine curve.
Define \(f\colon (0,1]\rightarrow \R\) as \(f(t)=\sin(1/t)\) for all \(t\in (0,1)\text{.}\) Let \(\Gamma_f=\{(t,f(t))\in \R^2\colon t\in (0,1)\}\text{,}\) the graph of \(f\text{.}\)
The set \(X=\overline{\Gamma_f}=\Gamma_f\cup \{0\}\times [0,1]\) is connected.
The set \(X\) is not path connected. More specifically, we can show that \((0,0)\) and \((1,\sin(1))\) are not connected by a path in \(X\text{.}\)
Proof.
Since \((0,1]\) is connected and \(f\colon (0,1]\rightarrow \R\) is continuous, its graph \(\Gamma_f\) is connected by Corollary 1.12.8. Since closure preserves connectedness, we conclude that \(X=\overline{\Gamma_f}\) is connected.
Suppose \(\phi\colon [0,1]\rightarrow X\) is a path connecting \((0,0)\) and \((1,\sin(1)\text{.}\) We can write \(\phi(t)=(\alpha(t), \beta(t))\) where the component functions \(x=\alpha(t), y=\beta(t)\) are both continuous. The set \(C=\{t\colon \alpha(t)=0\}=\alpha^{-1}(\{0\})\subseteq [0,1]\) is closed, since \(\alpha\) is continuous. Set \(t_0=\sup C=\max C\) (guaranteed to exist since \(C\subseteq[0,1]\) is closed), and let \(\phi(t_0)=(0,y_0)\text{.}\) Note that \(t_0\ne 1\) since \(\phi(1)=(1,\sin 1)\text{,}\) and that for all \(t> t_0\) we have \(\phi(t)\in \Gamma_t\) and hence \(y=\beta(t)=f(\alpha(t))\text{.}\) Take \(\epsilon=1\text{.}\) We will show that there is no \(\delta\)-ball around \(t_0\) such that \(\beta(B_\delta(t_0))\subseteq B_{1}(y_0)\text{,}\) contradicting the continuity of \(\beta\text{.}\)
For any any \(\delta> 0\text{,}\)\(\alpha([t_0, t_0+\delta))\) is an interval since \(\alpha(t)\) is continuous and \([t_0,t_0+\delta)\) is connected. This means that \(\alpha([t_0,t_0+\delta))\) contains an interval of the form \([0,c)\) for some \(c>0\text{.}\) Due to the highly oscillatory nature of \(f\text{,}\) we can find points \(x_1,x_2\in [0,c)\) satisfying \(f(x_1)=1, f(x_2)=-1\text{.}\) Since these points are in the image of \(\alpha\text{,}\) we have \(x_1=\alpha(t_1)\) and \(x_2=\alpha(t_2)\) for some \(t_1,t_2\in (t_0,t_0+\delta)\text{.}\) But then
Since \(d(\beta(t_1),\beta(t_2))=2\text{,}\) we cannot have \(\beta(t_1),\beta(t_2)\in B_{1}(y_0)\text{,}\) and thus \(\beta(B_\delta(t_0))\not\subseteq B_1(y_0)\text{,}\) as claimed.
Definition1.13.9.Connected components.
Let \(X\) be a topological space. The connected component of an element \(x\in X\) is the set \(C(x)\) defined as the union of all connected subsets of \(X\) containing \(x\text{.}\) Similarly, the path-connected component of $x$ is the set $C_p(x)$ defined as the union of all path connected subspaces of \(X\) containing \(x\text{.}\) A subspace \(A\subseteq X\) is maximally connected (resp., maximally path connected) if \(A\) is connected (resp. path connected), and if for all connected (resp. path connected) sets \(B\text{,}\) if \(A\subseteq B\text{,}\) then \(A=B\text{.}\)
Theorem1.13.10.Connected components.
Let \(X\) be a topological space.
For all \(x\in X\) the component \(C(x)\) is the unique maximally connected set containing \(x\text{,}\) and \(C_p(x)\) is the unique maximally path connected set containing \(x\text{.}\)
The collection \(\{C(x)\colon x\in X\}\) of all connected components is a partition of \(X\) by closed, maximally connected sets. In other words:
\(X=\bigcup_{x\in X}C(x)\text{.}\)
\(C(x)\ne C(y)\implies C(x)\cap C(y)=\emptyset\) for all \(x,y\in X\text{.}\)
\(C(x)\) is maximally connected for all \(x\text{.}\)
\(C(x)\) is closed for all \(x\text{.}\)
Similarly, the collection \(\{C_p(x)\colon x\in X\}\) of all path-connected components of \(X\) is a partition of \(X\) by maximally path-connected sets.
If \(A\) is connected (resp. path connected) and \(C\) is a connected component (resp. path connected component) of \(X\text{,}\) then either \(A\cap C=\emptyset\) or \(A\subseteq C\text{.}\)
For all \(x\in X\) we have \(C_p(x)\subseteq C(x)\text{,}\) and in fact \(\{C_p(y)\colon y\in C(x)\}\) is a partition of \(C(x)\) by path-connected components.
Proof.
Let us prove that \(C(x)\) is maximally connected. It is connected as it is the union of connected sets containing \(x\text{.}\) If \(C(x)\subseteq A\) and \(A\) is connected, then since \(x\in A\text{,}\) we have by definition \(A\subseteq C(x)\text{.}\) Thus \(C(x)\) is maximally connected. Lastly, if \(A\) is a maximally connected set containing \(x\text{,}\) then by definition we have \(A\subseteq C(x)\text{;}\) and since \(A\) is maximally connected we conclude that \(A=C(x)\text{.}\) This proves \(C(x)\) is the unique maximally connected set containing \(x\text{.}\)
The proof that \(C_p(x)\) is the unique maximally path-connected set containing \(x\) proceeds in exactly the same manner. We use the fact (not made official, but easy to show) that a union of path-path connected sets with a common intersection point is path connected.
It is clear that \(X=\bigcup_{x\in X}C(x)\text{,}\) since \(x\in C(x)\) for all \(x\in X\text{.}\) Furthermore, we know \(C(x)\) is maximally connected by \((1)\text{.}\) Since \(\overline{C(x)}\) is connected and \(C(x)\subseteq \overline{C(x)}\text{,}\) we must have \(C(x)=\overline{C(x)}\text{.}\) Thus \(C(x)\) is closed.
Lastly, to see that the components \(C(x)\) are disjoint, we will show that \(C(x)\cap C(y)\ne \emptyset\implies C(x)=C(y)\text{.}\) If \(C(x)\cap C(y)\ne \emptyset\text{,}\) then \(C(x)\cup C(y)\) is a connected set containing \(x\) and \(y\text{.}\) Since \(C(x)\) is maximally connected, we must have \(C(x)\cup C(y)=C(x)\text{,}\) and thus \(C(y)\subseteq C(x)\text{.}\) The same argument shows \(C(x)\subseteq C(y)\text{.}\) Thus \(C(x)=C(y)\text{.}\)
The proof here is exactly as with connected components. Again, we use the fact that the union of path-connected sets sharing a common point is path connected.
Suppose \(A\) is connected. If \(A\cap C\ne\emptyset\text{,}\) then \(A\cup C\) is connected by Theorem 1.12.6. Since \(C\) is maximally connected, we have \(A\cup C=C\text{.}\) Thus \(A\subseteq C\text{.}\) A similar argument applies to path-connected components.
Since \(C_p(x)\) is connected and \(C(x)\) is the union of all connected sets containing \(x\text{,}\) we have \(C_p(x)\subseteq C(x)\text{.}\) Since \(C(x)=C(y)\) for all \(y\in C(x)\text{,}\) we further have \(C_p(y)\subseteq C(x)\) for all \(y\in C(x)\text{.}\) It follows easily that \(C(x)=\bigcup_{y\in C(x)}C_p(y)\text{.}\) That this union is disjoint follows from (3).
