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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.

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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
    
See edited post. –  icurays1 Dec 14 '12 at 4:58

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.

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