4

$\mathbb{N}$ is the set of natural numbers.

Let $U_{\alpha \in A} \subset \mathbb{N}$ be the subset such that its complement $\mathbb{N}$ \ $U_\alpha$ is a finite subset.

Then $T= \{\emptyset, \mathbb{N}, U_\alpha |\alpha \in A \}$ is a topology on $\mathbb{N}$.

Prove that $\mathbb{N}$ is not path-connected space with the following definitions below

Definition: A topological space $X$ is path-connected space if for every points $a,b \in X$, there is a continuous map $f:[0,1]\rightarrow X$ such that $f(0)=a$ and $f(1)=b$.

Definition: Let $X$ and $Y$ be two topological spaces. A mapping $f:X \rightarrow Y$ is continuous at a point $a \in X$ if and only if for any open set $W$ containing $f(a)$ in $Y$, there exists an open set $G$ containing $a$ in $X$ such that $f(G) \subseteq W$.

Following statements are equivalent about continuity of map $f$ on two topological spaces $X$ and $Y$:

  • "$f$ is continuous on $X$" is equivalent to "for every open set $W$ in $Y$, the inverse image $f^{-1}(W)$ is open in $X$.
Ivo Terek
  • 77,665
MathsMy
  • 911

1 Answers1

2

U can easily show that:

$[0,1]\subset \mathbb R$ can not be written as countable union of closed and disjoint sets $\implies $ $\mathbb N$ with the cofinite topology is not patch-connected

Prove: If $\mathbb N$ where patch-connected then there is a continuous map $\gamma :[0,1] \rightarrow \mathbb N$ such that $\gamma(0)=x$ and $\gamma(1)=y$ for all x,y $\in \mathbb N$. Then $[0,1]=\gamma^{-1}(\mathbb N)=\gamma^{-1}(\bigcup_{i=1}^{\infty}\{i\})=\bigcup_{i=1}^{\infty}\gamma^{-1}\{i\}$ Since finite elements in the confinite topology are closed and $\gamma$ is continuous we have a countable union of closed disjoints sets. A contradiction.

Now it is enough to prove that there does not exist such an union.

Here is the basic idea: (Again by contradiction)

1) Suppose that $\bigcup_{i=1}^{\infty}B_n=[0,1]$ with closed and disjoint $B_n$

2) Construct a decreasing sequence of closed intervals $...I_4 \subset I_3 \subset I_2 \subset I_1 \subset [0,1]$ such that $B_n \cap I_n =\emptyset$

3) The set $\bigcap_{n=1}^{\infty}I_n$ is not empty. Let $x$ be an element of the set. Then $x$ is element of every $I_n$ and of exactly one $B_n$, hence $B_j \cap I_j=\{x\}$ for one $j\in \mathbb N$. A contradiction. The statement follows.

I left the construction of the intervals $I_n$ by the reader :)

Marm
  • 3,371
  • 1
  • 16
  • 24