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Are there a closed connected subspace $C$ of $\mathbb{R}^2$ and a continuous, bijective function $f:C\to\mathbb{R}^2$ that is not open?

If we remove the condition for $C$ to be connected, we have the function $f:\big([0,+\infty)\times\mathbb{R}\big)\cup\{(-1,0)\}\to\mathbb{R}^2$ defined by \begin{equation} f(x,y)= \begin{cases} &\bigg(e^y\cos\big(\frac{2\pi x}{x+1}\big),e^y\sin\big(\frac{2\pi x}{x+1}\big)\bigg)&,&(x,y)\neq (-1,0)\\ &(0,0)&,&(x,y)=(-1,0) \end{cases} \end{equation} which satisfy the rest of the conditions. See justification here.

But what is the answer if $C$ is closed and connected?

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1 Answer 1

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Are there a closed connected subspace $C$ of $\mathbb{R}^2$ and a continuous, bijective function $f:C\to\mathbb{R}^2$ that is not open?

No. Suppose $C\subseteq \mathbb R$ is closed and connected. Then either

(0) $C=\varnothing$,

(1) $C\simeq \{0\}$,

(2) $C\simeq [0,1]$,

(3) $C\simeq [0,\infty)$, or

(4) $C\simeq \mathbb R$.


(0) and (1): Cannot work because in this case $C$ and $\mathbb R^2$ have different cardinalities.

(2): Cannot work because $[0,1]$ is compact but $\mathbb R ^2$ is not.

(3): Suppose $f:[0,\infty)\to \mathbb R ^2$ is continuous and injective. Then for each $n\in\omega$, $f\restriction [0,n]$ is a homeomorphism, as a continuous function from a compact space to a Hausdorff space is closed. Thus $f[0,n]$ is closed and nowhere dense in $\mathbb R ^2$ (if it had interior then the image would have non-cut points, whereas $[0,n]$ has no cut points).Thus we find $f[0,\infty)$ is a countable union of closed nowhere dense subsets of $\mathbb R ^2$, which cannot be all of $\mathbb R ^2$ by the Baire Category Theorem for complete metric spaces. So $f$ is not surjective.

(4): Replace $[0,n]$ with $[-n,n]$ in the argument above.

EDIT: I now see that there was a typo in the title of the thread. I guess $C$ should be a subset of $\mathbb R ^2$ not $\mathbb R$. Nevertheless, I have ruled out some possibilities for $C$.

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