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I do not fully comprehend the definition of continuity in the topological way. For example:

Define $f:(\mathbb{R},T_{Eucl})\to(\mathbb{R},T_{Eucl})$ by $f(x)=\sqrt{x}$. I have a feeling that this function is not continous in the topological sense. Take $A:=(-1,1)$, then $f^{-1}{(A)}=[0,1)\notin T_{Eucl}$. So $f$ is not continuous. But this does not comply with results in Analysis? Or does it? Is this reasoning correct or is it not possible to look at $A$, because it is not contained in the image of $f$ (which is not states as a necessity in the definition)?

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    $f$ is not a function on $\mathbb{R}$ ;). – 2017-01-29
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    Obviously, thanks! Then this does not hold indeed. I was thinking about $x^2$ and $\sqrt{x}$ and mixed up. – 2017-01-29

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$f^{-1}(A)=(-1,1)$ not $[0,1)$ for $f(x)=x^2$, $A=(-1,1)$. And $\sqrt x$ is not defined on all $\mathbb R$. If we define it only on non-negative numbers $\mathbb R_+$, we see that the pre-image is $[0,1)$, but it is open in $\mathbb R_+$.

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The function $f(x) = \sqrt{x}$ is not a function from $\mathbb{R}$ to $\mathbb{R}$, as it is not defined for $x < 0$.

We can consider it as a function from $\mathbb{R}^+ = [0,\infty)$ to $\mathbb{R}$ but then we must consider the subspace topology on $\mathbb{R}^+$, which consists of all sets of the form $O \cap \mathbb{R}^+$ where $O$ is open in the Euclidean topology on $\mathbb{R}$.

NOw $$f^{-1}[(-1,1)] = [0,1) = (-1,1) \cap \mathbb{R}^+\text{,}$$

which is open in the subspace topology on $\mathbb{R}^+$. The definition works.