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?

  • 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


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 $\{ \emptyset , X\}$).


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