Suppose we are given a piecewise function $f: \mathbb{R} \backslash \{ 0 \} \to \mathbb{R}$, where $f(x)=1$ if $x>0$ and $f(x)=0$ if $x<0$.
This function will be continuous everywhere in "real analysis sense" (using $\epsilon$–$\delta$ definition) and in "topological sense" (using the standard topology and continuity defined there) because we excluded $ \{ 0 \}$ in the domain.
However, if we insist on including $\{ 0 \}$ in its domain, i.e. $f: \mathbb{R} \to \mathbb{R}$. This function will not be continuous everywhere in "real analysis sense" because it is undefined at $0$ in the first place. Since continuity in "topological sense" is equivalent to continuity in "real analysis sense", this function, with $0$ in the domain but undefined there, should also not be continuous in "topological sense".
My question is which among the subsets in the standard topology in the codomain does not have an open pre-image for this to be considered non-continuous in "topological sense"?
Best Answer
If we assign some value $f(0) =c$ for a $c \in \Bbb R$, then whatever $c$ is, the result will not be continuous with domain $\Bbb R$: for any neighbourhood $U$ of $0$, $f[U] = \{1,0,c\}$ and this can never lie entirely inside a neighbourhood of $f(0)$. (or, we could use that $f[U]$ should be connected for a continuous $f$, because $\Bbb R$ has a base of connected neighbourhoods). And as to preimages, take any open set in the codomain that includes $c$ but not $1$ or $0$, if $c \neq 0,1$, and then $f^{-1}[U]=\{0\}$ is not open. Or if $c=0$ we take $U= (-\frac12, \frac12)$ and then $f^{-1}[U] = (-\infty,0]$ which is not open ($0$ is not interior) or when $c=1$ we take $U=(\frac12,+\infty)$ which has $f^{-1}[U] = [0,+\infty)$, also not open. The local argument with connectedness works better IMHO.