9
$\begingroup$

Show that the Sierpinski space is a continuous image of $[0,1]$.

Consider the Sierpinski space $(S, \mathcal T)$ where $S=\{0,1\}$ and $\mathcal T= \{ \emptyset, S, \{1\}\}$.

Define $~f:[0,1]\to S$ by $$f(x) = \begin{cases}0 & x=0 \\ 1 & x \in (0,1]\end{cases}$$

Now $f^{-1}(\{1\})= (0,1]$, which is open in the relative topology of $[0,1]$ as a subset of $\mathbb R$.

and $f^{-1}(S)=[0,1]$, which is clearly open in $[0,1]$.

So we have found a continuous mapping $~f:I \to S$ and, consequently, $S$ is a continuous image of $[0,1]$.

Is this correct?

$\endgroup$
2
  • 4
    $\begingroup$ Yes, this is fine. $\endgroup$ Nov 14, 2016 at 0:29
  • $\begingroup$ Yes. The only non-empty open proper subset of $S$ is $\{1\} .$ So for continuity it is sufficient (& also necessary) that $f^{-1}\{1\}$ is open in $[0,1].$ For surjectivity it is sufficient (& also necessary) that $f([0,1])=S$. $\endgroup$ Jan 20, 2017 at 22:02

1 Answer 1

1
$\begingroup$

Yes. What you need to do is choose a non-empty proper open subset of $[0,1]$, map every point in it to $1$ and every point outside to $0$.

The interval can be replaced by any topological space $X$ whose topology is non-trivial (not reduced to $\{ \emptyset , X\}$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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