Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a topological space. We denote by $\tau_\mathbb{R}$ the final topology induced by the family of continuous maps $\varphi:\mathbb{R}\rightarrow X$, and by $\tau_I$ the final topology induced by the family of continuous maps $\psi:I\rightarrow X$, where $I$ is the closed unit interval $[0,1]$.

I want to show that $\tau_I=\tau_\mathbb{R}$.

The proof that $\tau_\mathbb{R}\subset \tau_I$ isn't too difficult:

Suppose $U$ is an element of $\tau_\mathbb{R}$. We wish to show that for every continuous map $\varphi:I\rightarrow X$ we have $\varphi^{-1}(U)$ is open. To achieve this, we extend $\varphi:I\rightarrow X$ to a map $\overline{\varphi}:\mathbb{R}\rightarrow X$ by defining $$\overline{\varphi}(x)=\begin{cases} \varphi(x) & x\in I \\ \varphi(1) & x>1 \\ \varphi(0) & x<0 \end{cases}$$ which is clearly continuous. Thus, since $U$ is an element of $\tau_\mathbb{R}$, we have $\overline{\varphi}^{-1}(U)$ is open, by which $I\cap \overline{\varphi}^{-1}(U)=\varphi^{-1}(U)$ is open in $I$.

However, I can't think of any good approach to take in showing that $\tau_I\subset \tau_\mathbb{R}$. Any help would be greatly appreciated.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Let $f:\mathbb{R}\to X$ be continuous, and take $U\in\tau_I$.

To show that $f^{-1}(U)$ is open, it is sufficient to show that $f^{-1}(U)\cap J$ is open in $J$ for every closed interval $J\subset\mathbb{R}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.