0
$\begingroup$

There must exist a continuous surjective function $f$ from $(0,1)$ to $[0,1]$. Since $f$ is continuous and $[0,1]$ is closed, then $f^{-1}([0,1])$ is closed either. However, $f^{-1}([0,1])=(0,1)$, which is open in $\mathbb{R}$.

How should I understand the statement that "the preimage of a closed set is closed"?

$\endgroup$
  • $\begingroup$ $(0,1)$ is open in $\mathbb{R}$, sure. But it's closed in the subspace topology on $(0,1)$, which is the relevant space. $\endgroup$ – Patrick Stevens Mar 29 '17 at 7:52
  • $\begingroup$ The domain of $f$ is $(0,1)$, not $\mathbb{R}$, and $(0,1)$ is certainly closed in itself. $\endgroup$ – quasi Mar 29 '17 at 7:53
  • $\begingroup$ For an example of a continuous surjective function from $(0,1)$ to $[0,1],\;$ let $$f\;{:}\;(0,1) \to [0,1]\;\;\text{be defined by}\;\;f(x) = \sin{2\pi x}$$ $\endgroup$ – quasi Mar 29 '17 at 8:06
1
$\begingroup$

To talk about a continuous function, we need to specify the domain, the codomain, and the topology on each of them. In particular, if the domain is a subset of the reals, we need to treat that subset as a topological space of its own rather than as a subset. As the whole space, it is certainly closed.

Note that if your function (that takes (0,1) to [0,1]) is defined continuously on a larger subset of the reals that includes 0 and 1, then the preimage of [0,1] will indeed also include 0 and 1.

$\endgroup$

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.