Is this definition of local continuity (with open sets) accurate? I am looking for a definition of local continuity (continuity at a point) with open sets.
A quick google search has led me to this but I could not understand the fourth slide...
But before the google search, I had invented up a little definition that could be reasonnably true. It is the following:
Let's pick a function $f$ and a point $(x_0,y_0)$ of the function.

If some set in the domain is:

*

*the preimage (by the function $f$) of an open interval (in the codomain) containing the ordinate of the point ($y_0$)

*a set which contains the $(x_0,0)$ point

*open



THEN, the function $f$ is continuous at the point $(x_0,y_0)$

What do you think of this local continuity definition? Is it correct?
 A: Your notations are a little woobly:

*

*In the second bullet, your set $f^{-1}(I)\subset\Bbb R$ must containt $x_0,$ not $(x_0,0).$

*$f$ will be continuous at the point $x_0,$ not $(x_0,y_0).$

*$f^{-1}(I)$ itself may not be open.

But the more important point is the lack of quantifiers. Every function $f:\Bbb R\to\Bbb R$ s.t. $f(x_0)=y_0$ satisfies your condition, without necessarily being continuous at $x_0.$ Here is a sensible formulation:

For every real open interval $I\ni y_0,$ the preimage $f^{-1}(I)$ contains some open set containing $x_0.$

However, this is the usual definition of continuity at $x_0.$ The notion of local continuity at $x_0$ defined in the fifth slide of your link, which I didn't know before, is (equivalent to) a slight modification of the previous definition:

There exists an open interval $J\ni x_0$ such that $x_0\in\bar J$ and for every real open interval $I\ni y_0,$ the preimage $f^{-1}(I)$ contains the intersection of $J$ with some open set containing $x_0.$

