# Prove there isn't a continuous surjection $f: [0, 1] \to \Bbb R$ (without compactness)

Definitions:

Continuous: A map $f: X \to Y$, where $X$ and $Y$ are topological spaces, is continuous if the preimage in $X$ of any open set in $Y$ is open.

Subspace topology: If $(X, \mathcal{T})$ is a topological space, the subspace topology on a set $S \subset X$ is $\mathcal{T}_S = \{S \cap U : U \in \mathcal{T}\}$.

The problem says that $[0, 1]$ is a topological space with the subspace topology, meaning some sets that I would not think is open, like $[0, 1]$, is an open set, so this kind of threw me off. I know from a different class that continuity preserves compactness, so I already know that there doesn't exist a continuous function, but we're not allowed to use compactness for this problem.

Is there a way to show using the subspace topology of $[0, 1]$ that there isn't a continuous surjection from the topological space $[0, 1]$ with the subspace topology onto $\Bbb R$?

• Hint: The image of a connected subset is connected. Now what happens to the image of $[0,1)$? – Tobias Kildetoft Apr 29 '15 at 12:04
• You may not even use that a continuous function on $[a,b]$ assumes its maximum? – Hagen von Eitzen Apr 29 '15 at 12:05
• Think of it this way then: Let $x$ be the image of $1$ and split the complement of $\{x\}$ into two disjoint open sets in the obvious way. The preimages of these sets are now disjoint and open. – Tobias Kildetoft Apr 29 '15 at 12:24
• @TobiasKildetoft The preimages of disjoint sets need not be disjoint – Hagen von Eitzen Apr 29 '15 at 12:30
• It might not be all of $[0,1)$ since other elements might be sent to $x$ also. – Tobias Kildetoft Apr 29 '15 at 13:59

Let's cheat and only use that $[0,1]$ is closed and bounded:
For each $k\in\mathbb N$ pick $x_k\in[0,1]$ with $f(x_k)=k$. Starting with $I_0=[0,1]$, which contains all $x_k$, we can repeatedly split $I_n=[a_n,b_n]$ into two subintervals $[a_n,\frac{a_n+b_n}2]$, $[\frac{a_n+b_n}2,b]$ and one of these contains infinitely many $x_k$ and we let $I_{n+1}$ be that interval. Then the intersection $\bigcap_{n\in\mathbb N} I_n$ is a singleton set $\{a\}$, where $a\in[0,1]$. A suitable subsequence $x_{k_n}$ of the $x_k$ converges to $a$, henc by continuity $f(x_{k_n})\to f(a)$, which is absurd.
• @eyes: I would be surprised if you don't have "bounded" on the real line. A bounded subset of $\mathbb{R}$ simply means a subset contained in $[-a,+a]$ for some real number $a>0$. – Lee Mosher Apr 29 '15 at 20:34