So I am beginning Munkres' textbook on topology.
The topological definition of continuity reads:
$f:X\rightarrow Y$ is continuous if for each open subset $V\subset Y$, $f^{-1}(V)$ is an open subset of $X$.
Of course, it does fit the epsilon-delta definition of continuity since in
$\forall\epsilon\exists\delta,|x-x_0|<\delta\Rightarrow|f(x)-f(x_0)|<\epsilon$
$|x-x_0|<\delta$ and $|f(x)-f(x_0)|<\epsilon$ are both open.
Also, since openess of a set means closeness of that set's complement, the following epsilon-delta definition (which is valid) also agree with the topological definition:
$\forall\epsilon\exists\delta,|x-x_0|\le\delta\Rightarrow|f(x)-f(x_0)|\le\epsilon$
Yet my question is, how about the following epsilon-delta definition?
$\forall\epsilon\exists\delta,|x-x_0|\le\delta\Rightarrow|f(x)-f(x_0)|<\epsilon$
1.) Is it a valid definition of continuity of $\mathbb{R}^n$ regardless of the topological definition?
2.) If yes, does it agree with the topological definition of continuity?