TITLE: Proof with Topological Definition of Continuity
QUESTION [0 upvotes]: I am using the following definition of continuity: $f:X \rightarrow Y$ is continuous if for each open subset $V\subset Y$, $f^{−1}(V)$ is an open subset of $X$.
Using this definition, how would I show that $f(x) = \sqrt{x}$ is continuous? If I choose an interval $(a, b),$ I get that $f^{-1}(a, b) = (a^2, b^2),$ which is open. But how do I show that $f$ is continuous at $0$, as I cannot construct an open interval containing $0$ because negative numbers are not in the range of $f$?

REPLY [1 votes]: Given a topological space $(X,\mu)$ and a subset $Y\subset X$, we define
$\sigma=\{A\cap Y: A\in\mu\}$ as the induced topology in $Y$ and we obtain a new topological space $(Y,\sigma)$ (note that the open sets in $Y$, element of $\sigma$ couldn't be open sets in $X$ -- depends if $Y$ is or not is an open in $X$).
For example, if $X=\Bbb R$ with the canonical topology and $Y=[0,1)$, the set $[0,\frac12)$ is open in $Y$ (because $[0,\frac12) = (-1,\frac12)\cap Y$ and $(-1,\frac12)$ is open in $X$).
Therefore, you have to think about where $f$ is defined.
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