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In topology, continuity is defined as:

A function $f:X\rightarrow Y$ is continuous if the inverse image of an open set in $Y$ is an open set in $X$.

I have a problem to use it to check the non-continuous function. For example, in J.Munkres' book Topology (2nd Edition) (Pg.109), there is an example $$f(x)=\begin{cases} x-2, & x<0\\ x+2, & x\geq0 \end{cases}$$ The domain of this function (i.e. the $X$ in the definition) is $\mathbb{R}$, i.e. $x\in(-\infty,\infty)$; the codomain (i.e. the $Y$ in the definition) is $(-\infty,-2)\cup[2,\infty)$. To prove this function is discontinuous at $x=0$, the book choose the open set of $f$ as $(1,3)$, and thus the inverse image is $[0,1)$, which is not an open set.

The problem is:

the open set $(1,3)$ is not the subset of the codomain $(-\infty,-2)\cup[2,\infty)$. How can we choose it?

In my point of view, within the codomain $(-\infty,-2)\cup[2,\infty)$, all allowed open sets are either within the subset $(-\infty,-2)$ or within the subset $[2,\infty)$. Within these two parts, the function are always continuous. How can we prove the discontinuous?

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    $\begingroup$ You can, if no codomain is actually specified; it just makes things unnecessarily complicated. If you treat $Y=(\leftarrow,-2)\cup[2,\to)$ as the codomain, then you have to understand that it inherits its topology from $\Bbb R$, so $(1,3)\cap Y=[2,3)$ is an open subset of $Y$. $\endgroup$ Commented May 31, 2020 at 22:30
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    $\begingroup$ But $Y$ is an open set in the subspace $Y$ of $\Bbb R$. Recall that the subspace topology on $Y$ is $\{U\cap Y:U\text{ is open in }\Bbb R\}$. $\endgroup$ Commented May 31, 2020 at 22:40
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    $\begingroup$ If you insist on using $Y$ as the codomain instead of the natural one, $\Bbb R$, then you have to use $[2,3)$ as your open set in the codomain, not $(1,3)$. In practical terms it comes to exactly the same thing. $\endgroup$ Commented May 31, 2020 at 22:45
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    $\begingroup$ $[2,3)$ is an open set in $Y$ when $Y$ has the subspace topology that it inherits from $\Bbb R$. If you take $Y$ as the codomain, then you must work in that topology. If you want to use the topology of $\Bbb R$ itself, you must take $\Bbb R$ as the codomain. $\endgroup$ Commented May 31, 2020 at 22:54
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    $\begingroup$ Of course there are topologies on $Y$ in which $[2,3)$ is a closed set; the discrete topology is one of them. But none of them is at all relevant to the original problem. In that problem you are expected to understand that $f$ is a function from $\Bbb R$ to $\Bbb R$. If you artificially take its range $Y$ to be its codomain, then you are still required to use the relevant topology, which is the subspace topology that $Y$ inherits from $\Bbb R$. $\endgroup$ Commented May 31, 2020 at 22:58

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The codomain of your function is $\Bbb R$, not $(-\infty,-2)\cup[2,\infty)$. So, there is no problem, since $(1,3)\subset\Bbb R$.

If you want to see $f$ as a map from $\Bbb R$ into $Y=(-\infty,-2)\cup[2,\infty)$, then, instead of $(1,3)$, take its intersection with$Y$, which is $[2,3)$. It is an open subset of $Y$, but $f^{-1}\bigl([2,3)\bigr)$ is not an open subset of $\Bbb R$.

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  • $\begingroup$ I don't get it clearly. If we talk the intersection within $\mathbb{R}$, $Y=(-\infty, -2)\cup[2,\infty)$ itself is not an open set in $\mathbb{R}$, how can the intersection to be an open set? If we talk the intersection within $Y$ (then $Y$ is an open set), but the $(1,3)$ is out of bounds (is not within $Y$), how can we make this intersection? $\endgroup$ Commented May 31, 2020 at 22:44
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    $\begingroup$ By definition, a subset $A$ of $Y$ is an open set if there is an open subset $A^\star$ of $\Bbb R$ such that $A=Y\cap A^\star$. And $[2,3)=Y\cap(1,3)$. $\endgroup$ Commented May 31, 2020 at 22:48
  • $\begingroup$ Can you say it a little more clear? This is the definition of inherited topology? I think we can definitely treat $Y$ isolated to a bigger set. And define a topology with $[2,3)$ a close set? $\endgroup$ Commented May 31, 2020 at 22:58
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    $\begingroup$ Yes, what I wrote is the definition of inherited topology (although the usual name is subspace topology). $\endgroup$ Commented May 31, 2020 at 23:00

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