According to Folland's Real Analysis, the following definitions of continuity at a point are equivalent:
Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces. A function $f \colon X \to Y$ is said to be continuous at a point $x \in X$ if, and only if, for every neighborhood $O$ of $f(x)$ it holds that $f^{-1}(O)$ is a neighborhood of $x$.
Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces. A function $f \colon X \to Y$ is said to be continuous at a point $x \in X$ if, and only if, for every neighborhood $O$ of $f(x)$ there is a neighborhood $U$ of $x$ satisfying $f(U) \subseteq O$.
Folland's definition of a neighborhood of $x$ is a set $A$ such that $x \in$ int$(A)$. I've been trying to prove it using a different definition (a neighborhood is an open set containing the point). Does the result still hold under this different definition? Regardless of the answer to this, how can I show it (preferably with my definition, but I guess the other one would already shine some light)? That the first implies the second seemed clear (pick $U = f^{-1}(O)$), but I'm not being able to prove the inverse.
Best Answer
Let $(1)$ and $(2)$ denote Folland's two definitions, and $(1')$ and $(2')$ denote these definitions where 'neighborhood' is replaced by 'open neighborhood'. Then $(1)$, $(2)$ and $(2')$ are all equivalent but $(1')$ is not equivalent to the others.
To see that $(1')$ and $(2')$ are not equivalent, define $f:\mathbb{R} \to \mathbb{R}$ by $$f(x) = \left\{ \begin{array}{ll} 0 & \mbox{if $x < 0$}\\ 1 \quad &\mbox{if $x \geq 0$.}\end{array} \right.$$
We will show that $(1')$ and $(2')$ disagree about whether $f$ is continuous at $1$.
Using $(1')$: The interval $O = (\frac12,\frac32)$ is an open neighborhood of $f(1) = 1$ but $f^{-1}(O) = [0,\infty)$ is not open, so $f$ is not continuous at $1$.
Using $(2')$: Let $O$ be a open neighborhood of $f(1) = 1$. Let $U = (0,2)$ which is an open neighborhood of $1$. Then $f(U) = \{1\} \subseteq O$, so $f$ is continuous at $1$.