1

I want to show that $\mathbb{R}$ is connected given that the unit interval is connected. I only know the definition of connectedness (a space $X$ is connected if the only separations of it are the trivial ones ) and I know that the \sim -equivalence class of $x$ in $X$ is connected where $x \sim y$ iff there is a connected subspace $C \subseteq X$ s.t. $x,y \in C$. Also, I know that the definition of trivial separation is that if $X = U \bigcup V$ then $X=U$ or $X=V.$ and a separation of a space $X$ in general means $X= U \bigcup V$ where $U \bigcap V = \emptyset,$ and $U,V$ are both open in $X.$

My Question is:

Knowing all the above definitions, still I do not know how to prove that $\mathbb{R}$ is connected through them, could anyone help me in proving this by the above tools?

I know that there is this question here Showing that $\mathbb{R}$ is connected but this does not prove the statement by the tools I want.

José Carlos Santos
  • 440,053
Intuition
  • 3,043
  • 1
    The open or the closed unit interval? If it's the open: Take some continuous bijective function $f : (0, 1) \to \mathbb{R}$ (use the tangent function) and then observe that a eparation of $\mathbb{R}$ yields a separation of $(0, 1)$ by taking preimages. –  Feb 14 '20 at 13:58
  • @PaulK the closed unit interval – Intuition Feb 14 '20 at 14:03

2 Answers2

5

Suppose that $\mathbb R$ is not connected. Then, as you know, you can write $\mathbb R$ as $U\cup V$, where $U\cap V=\emptyset$ and $U$ and $V$ are non-empty open subsets of $\mathbb R$. Take $a\in U$ and $b\in V$. Since $U\cap V=\emptyset$, $a\neq b$. I will assume that $a<b$. Then $[a,b]$ is disconnected, because $[a,b]\cap U$ and $[a,b]\cap V$ are non-empty open subsets of $[a,b]$ whose intersection is empty and whose union is $[a,b]$. But this is impossibile, since $[a,b]$ and $[0,1]$ are homeomorphic (consider the map from $[0,1]$ into $[a,b]$ defined by $x\mapsto a+x(b-a)$) and $[0,1]$ is connected.

José Carlos Santos
  • 440,053
2

There are multiple ways to do that, e.g.

  1. If "unit interval" $I$ means either $(0,1)$ or $[0,1)$ or $(0,1]$ then there is a continuous surjection $f:I\to\mathbb{R}$. For $(0,1)$ it is a variant of $\tan(x)$, for two others a variant of $x\sin(\tan(x))$. But this approach does not work for $[0,1]$.

  2. Another way is to realize that if $\{C_i\}$ is a collection of connected subsets such that $\bigcap C_i\neq\emptyset$ then $\bigcup C_i$ is connected (see here). With that you can take the collection $\{[-t,t]\}_{t\geq 1}$.

freakish
  • 47,446