# Is this mapping continuous?

If there is a mapping that is closed and open, is that enough to claim that that mapping is continuous? I can't really prove that or disprove that.

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No. Example: the floor function $x\mapsto\lfloor x\rfloor$, from $(\Bbb{R},\vert\cdot\vert)$ to $\Bbb{Z}$ (with the discrete topology) is both open and closed, but the preimage of $\{1\}$ (an open set under the discrete topology) is $[1,2)$, which is not open under $\vert\cdot\vert$.
In fact any map to a space endowed with the discrete topology will always be both open and closed. Continuity is still possible though, just not guaranteed. For example a constant function like $f:(\Bbb{R},\vert\cdot\vert)\rightarrow(\Bbb{R},2^\Bbb{R})$ with $x\mapsto 1$; the preimage of any set containing $1$ is $\Bbb{R}$, which is open, while the preimage of any set not containing $1$ is empty, also open. So this map is continuous.
so is that because discrete metric space always closed and open so its mapping can be both closed and open but it can't be continuous as for $\epsilon <1$ there doesn't exist any $x$ such that $|f(x)-f(x_0)|<1$ whenever $|x-x_0|<\delta$ – Mathematics Dec 14 '12 at 4:51
Take any nonempty set $X$, and let $(X, d)$ be the discrete space and $(X, \tau)$ the indiscrete space. The identity from the latter to the former is both closed and open, but not continuous.