Random math: 37

Source of most content: Gunhee Cho, Lecture note by Matsumura and Topology by Munkres.


Def. Given T.S. $(X, T)$, $X$ is path-connected if $\forall p, q \in X$, $\exists$ cts map $f:[a, b] (\subseteq \mathbb{R}) \to X$ s.t. $f(a)=p$ and $f(q)=b$. We call $f$ a path.

NB1; If $X$ is path-connected, then $X$ is connected.

Suppose $X$ is disconnected, i.e., $X = U \mathbin{\unicode{x228D}} V$ for non-empty open sets $U, V$. Since $f$ is continuous and $[a, b]$ is connected (NB6), $f([a, b])=X$ is also connected (2) which is a contradiction.

NB2; Every ball is path-connected.

Let $B_{\epsilon}(0) \subsetneq \mathbb{R}^n$ be a $\epsilon$-ball centered zero. Then, $\forall x,y \in B_{\epsilon}(0) = B$, $\forall t \in [0, 1]$, let $f(t) := tx + (1-t)y$. Since $\Vert x \Vert, \Vert y \Vert \leq \epsilon$, $\Vert f(t) \Vert \leq \epsilon$, i.e., $f(t) \in B$.

NB3; The closure of a connected space is connected, but the closure of a path connected space may not be path connected.
NB4; $S^{n-1} = \{ (x_1, \ldots, x_n) | x_1^2 + \ldots + x_n^2 = 1, x_i \in \mathbb{R} \} \subsetneq \mathbb{R}^n$. Consider $f : \mathbb{R}^n \setminus \{0\} \to S^{n-1} (x \mapsto \frac{x}{\Vert x \Vert}). $
NB5; Example taht is connected but not path-connected. (Topologist’s sine curve)

Consider $I= [0, 1]$ which is connected. Then, $I \times I$ is also connected. If $I \times I$ is path-connected, we can tak $(0,0)$ and $(1,1)$. Then, $\exists$ cts $f: [a, b] \to I \times I$ with $f(a) = (0,0)$ and $f(b)=(1,1)$. By IVT, $\forall x \in I$, $f^{-1}(x \times (0,1)) \ne \emptyset$. Note that $\forall x \ne y$ in $I$, $f^{-1}(x\times (0,1)) \cap f^{-1}(y \times (0,1))=\emptyset$. Since $f^{-1}(x\times (0,1))$ is open in $[a, b]$, it contains some interval in $f^{-1}(x\times (0,1))$. Then, by the density of $\mathbb{Q}$, choose $q_x$ in that interval. In this way, we can define $\Phi : I \to \mathbb{Q} (x \mapsto q_x)$. Then, $\Phi$ is injective by construction. This implies that $|I| \leq |Q|$, which is a contradiction since $I$ is uncountable.

Locally connected

Def. Given T.S. $(X, T)$, $X$ is locally connected if $\forall x \in X,$ $\forall$ nbhd $U_x$ of $x$, $\exists$ connected nbhd $V_x$ of $x$ s.t. $x \in V_x \subseteq U_x$.

Def. Given T.S. $(X, T)$, $X$ is locally path-connected if $\forall x \in X,$ $\forall$ nbhd $U_x$ of $x$, $\exists$ path-connected nbhd $V_x$ of $x$ s.t. $x \in V_x \subseteq U_x$.

NB6; $\mathbb{R}-\{0\}$ is not connected but locaaly connected.
NB7; $\mathbb{Q}$ is not connected and not locally connected.