3
$\begingroup$

I know there are many examples when the domain and co-domain do not coincide. Taking the identity on $X$ from $(X,\tau_1)$ to $(X,\tau_2)$ when $\tau_2$ is coarser than $\tau_1$ gives an infinite family of examples.

However I have been struggling to find an example of such a function between the same topology. I came to think about this because in $\mathbb R$ any bijective continuous function is a homeomorphism.

Thank you in advance, regards.

$\endgroup$
1
$\begingroup$

Let $K=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the topology that it inherits from $\Bbb R$, let $\Bbb N$ have the discrete topology, and let $X=\Bbb N\times K$ with the product topology. Let

$$A=\{\langle n,y\rangle\in X:n\text{ is odd}\}\;;$$

$A$ is countably infinite, so let $\varphi:\Bbb N\times\{1\}\to A$ be any bijection. Then

$$f:X\to X:\langle n,y\rangle\mapsto\begin{cases} \langle 2n,y\rangle,&\text{if }y=0\\ \left\langle 2n,\frac1{k-1}\right\rangle,&\text{if }y=\frac1k<1\\ \varphi(\langle n,1\rangle),&\text{if }y=1 \end{cases}$$

is a continuous bijection that is not open: $f$ maps the isolated point $\varphi^{-1}(\langle 0,0\rangle)$ to the non-isolated point $\langle 0,0\rangle$.

This is simpler than it may at first appear. Pictorially, $X$ is just the disjoint union of countably infinitely many simple sequences:

$$\begin{array}{c|cc} 0&\color{brown}\bullet&\color{brown}\bullet&\color{brown}\bullet&\color{brown}\bullet&\ldots\\ \uparrow&\uparrow&\uparrow&\uparrow&\uparrow&\cdots\\ \frac14&\bullet&\bullet&\bullet&\bullet&\cdots\\ \frac13&\bullet&\bullet&\bullet&\bullet&\cdots\\ \frac12&\bullet&\bullet&\bullet&\bullet&\cdots\\ 1&\bullet&\bullet&\bullet&\bullet&\cdots\\ \hline &0&1&2&3&\cdots \end{array}$$

$A$ consists of the odd-numbered columns. The map $f$ takes all of column $n$ except the bottom point to column $2n$, moving the isolated (black) points down one but leaving the limit (brown) points at the top; the (blue) point at the bottom of column $1$ goes somewhere in $A$.

$$\begin{array}{c|cc} 0&\color{brown}\bullet&\rightarrow&\color{brown}\bullet\\ &\uparrow&&\uparrow&\\ \frac14&\bullet&&\bullet\\ &&\searrow\\ \frac13&\bullet&&\bullet\\ &&\searrow\\ \frac12&\bullet&&\bullet\\ &&\searrow\\ 1&\color{blue}\bullet&&\bullet\\ \hline &1&&2 \end{array}$$

This maps all of $X$ except the bottom row bijectively (and homeomorphically) onto $X\setminus A$, the even columns. The points in the bottom row are isolated, so a continuous map can send them anywhere; $\varphi$ (and hence $f$) simply sends them bijectively to $A$.

$\endgroup$
3
$\begingroup$

The simplest - in the sense of least things to verify - examples that I can think of are linear operators on some infinite-dimensional vector spaces. For example, endow the space

$$c_{00}(\mathbb{N}) := \bigl\{ x \colon \mathbb{N}\to \mathbb{C} : \bigl(\exists k\in\mathbb{N}\bigr)(n \geqslant k \implies x(n) = 0)\bigr\}$$

with an $\ell^p$-norm for $1 \leqslant p \leqslant +\infty$, and consider the operator $T$ given by

$$T(x)(n) = 2^{-n}\cdot x(n).$$

Then $T$ is linear, continuous and bijective (easily checked), but its inverse is unbounded, hence not continuous.

$\endgroup$
2
$\begingroup$

Let $\tau$ be the usual topology of $\mathbb R.$ Define a finer topology $\tau_1$ as follows: $$\tau_1=\{X\subseteq\mathbb R:X\cap(0,\infty)\in\tau\}.$$ The function $f(x)=x+1$ is a continuous bijection, but not a homeomorphism, from $(\mathbb R,\tau_1)$ to itself.

$\endgroup$

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.