Topologies and continuity: continuous iff continuous at every point

Question

Following Pedersen's Analysis Now, it seems that involving topologies in the definition of continuous functions, yet ignoring them in defining point-wise continuity, makes it difficult (impossible?) to show equivalence between the two.

Pedersen's definitions

Let (X, $\tau$) and (Y, $\sigma$) be topological spaces. A function $f: X \implies Y$ is said to be continuous if $f^{-1}(A) \in \tau$ for every $A$ in $\sigma$. It is said to be continuous at a point in $X$ if $f^{-1} \in O(x)$ for every $A$ in $O(f(x))$. [Where $O(x)$ is the neighborhood filter: the system of neighborhoods about point $x$. A subset is a neighborhood of $x$ if it contains an open set which contains $x$.]

Pedersen's proof for continuous iff continuous at every point, reverse direction:

If $f$ is continuous at every point and $A \in \sigma$, take $x$ in $f^{-1}(A)$. Thus $A \in O(f(x))$, whence $f^{-1}(A) \in O(x)$; so that $f^{-1}(A)$ is a neighborhood of every point it contains and, consequently, is open.

That doesn't seem sufficient to me. In the trivial example, assume the topology $\tau$ contains only the sets $\emptyset$ and $X$, while $\sigma$ is a more robust topology. Then we could have many $A \in \sigma$ for which, while $f^{-1}(A)$ is open, $f^{-1}(A) \not\in \tau$ (although $f^{-1}(A)$ would be a subset of a set in $\tau$).

What am I misunderstanding here?

Answer 1

What you are missing here is a contradiction in what you wrote. You cannot say that $f^{-1}(A)$ is open and, at the same time, say that $f^{-1}(A)\notin\tau$ since, by definition, the open subsets of $X$ in the topological space $(X,\tau)$ are the elements of $\tau$.