Hopefully, most people reading this are familiar with the notion of continuity in some format. However, this definition is a very curious exploration in which via something seemingly abstract we are able to extrapolate the one of the most basic practical concepts in calculus, topology and mathematics in general. But before we get into that, we need to learn about open sets (not same as open intervals).

## Open Sets

A subset $O$ of $ℝ$ is open if for each point $x \in O$ there exists an open interval $(a, b)$ that contains $x$ and is contained in $O$ .

Examples :

- $ℝ$ is an open set.
- An open interval is an open set.
- Intersection of two open sets is an open set.

and so on.

## Continuity

Definition : A function $f : ℝ→ℝ$ is continuous if for each open set $O$ in $ℝ$ the inverse image $f^{-1} (O) = {x ∈ ℝ | f (x) ∈ O}$ is also an open set.

Now, lets see how this corresponds to the usual definition of continuity hopefully all of us are familiar with.

Consider what would happen if the condition wad not met for some function $f$. So, there is an open set $O$ for which $f^{-1} (O)$ is not open. Thus, $∃ x_0 ∈ f^{-1}(O)$ for which $x_0$ is not contained in any arbitrary open interval $(a,b)$ which is further contained in $f^{-1}(O)$.

Intuitively, think of this as if $x_0$ is at the edge of $f^{-1} (O)$ and the complement of $f^{-1} (O)$.

As $x_0$ is supposed to at the “edge” of $f^{-1} (O)$ and the complement of $f^{-1} (O)$, we can say that there will be points, say $x$, which will be arbitrarily close to $x_0$ for which $x ∈ \text{complement of} ^{-1} (O)$.

Thus,

$$x ∉ f^{-1} (O), \

f(x) ∉ O$$

However,

$$x_0 ∈ f^{-1}(O), \

f (x_0) ∈ O$$

Which implies that : $$f (x_0) ∈ (c,d)\, \text{where} \ (c,d) ∈ O $$

Intuitively, think of this the same way you did when $x_0$ was at the edge of $f^{-1} (O)$. However, in this case, $f (x_0)$ will be in the interior of $O$ and not on its edge. Whereas, $f(x) ∉ O$ will be in the exterior of $O$. Hence, there is always a finite positive distance between $f (x_0)$ and $f(x)$ which will comprise of the edge of the $O$.

But as $x$ can be arbitrarily close to $x_0$, $x$ can be considered to be extremely close to $x_0$. Yet there would exist a finite distance between $f(x)$ and $f (x_0)$.