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Find example of a mapping that is open but not continuous and a mapping that is closed but not continuous.

I striving with these questions.

I thought of using for the first case a map between $(\mathbb{Z},\tau)$ where $\tau$ is the co-finite topology to $(\mathbb{R},\tau')$ with the standard topology. I think the mapping would not be continuous and since any open set in $\mathbb{R}$ is infinite and uncountable and the inverse image would need to be countable. However I am not seeing how to prove this with more rigorous mathematical terminology and I do not see how this mapping might be open.

Question:

Can someone help me solve this question?

Thanks in advance!

Pedro Gomes
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  • The inclusion $f:\mathbb{Z}\rightarrow\mathbb{R}$, $f(t)=t$ is not continuous because $f^{-1}(0,1)={0}$ and ${0}$ is not open in the co-finite topology. On the other hand $f$ is also not open because the image of $\mathbb{Z}$ is not open in $\mathbb{R}$. – Yanko Dec 15 '18 at 17:41
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    Identity from from the real with it's usual topology to the reals with the discrete topology is open and closed while not being continuous. – Robert Thingum Dec 16 '18 at 01:54

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