A continuous function is not necessarily an open map

A continuous function $ f: X \to Y $ does not necessarily map open sets in $ X $ to open sets in $ Y $.

Continuity does not imply that open sets are preserved, unlike with open maps.

Therefore, a continuous function is not necessarily an open map.

What is an open map? An open map $ f: X \to Y $ maps every open set in $ X $ to an open set in $ Y $.

In other words, not all continuous functions are open. Even if a function is continuous, it doesn't mean that the image of an open set in the domain will be open in the codomain.

A practical example
Difference between continuous and open maps

A practical example

Consider the function $ f(x) = x^2 $, which is continuous on $ \mathbb{R} $.

Take the open set $ (-2, 2) $ in $ \mathbb{R} $, which includes all real numbers between $ -2 $ and $ 2 $.

Now apply the function $ f = x^2 $ to this set.

$$ f(-2) = (-2)^2 = 4 \\ f(0) = 0^2 = 0 \\ f(2) = 2^2 = 4 $$

The image of $ (-2, 2) $ under $ f(x) = x^2 $ is the interval $ [0, 4) $, which is not an open set.

In fact, $ 0 $ is in the interval, but there is no neighborhood of $ 0 $ entirely contained within $ [0, 4) $, since $ 0 $ is a closed lower bound.

This shows that, even though $ f(x) = x^2 $ is a continuous function, it does not map the open set $ (-2, 2) $ to an open set, as expected.

Thus, while $ f(x) = x^2 $ is continuous at every point in its domain, it is not an open map.

Difference between continuous and open maps

The concepts of continuity and openness differ because they preserve open sets in distinct ways.

Continuous function (in terms of open sets)
A function $ f: X \to Y $ is continuous if the preimage of every open set in $ Y $ is open in $ X $. In other words, if you take an open set $ U $ in the codomain $ Y $, its preimage $ f^{-1}(U) $ must be open in the domain $ X $.
Continuity ensures that pulling open sets back from the codomain $ Y $ through $ f $ always results in open sets in the domain $ X $. Continuity concerns itself with how open sets in the codomain behave when "pulled back" to the domain, without any regard for what happens when open sets in the domain are "pushed forward" to the codomain.
Open maps (open function)
A function $ f: X \to Y $ is open if it maps open sets in $ X $ to open sets in $ Y $. In other words, if you take an open set $ V $ in the domain $ X $, its image $ f(V) $ must be an open set in $ Y $.
Openness focuses on mapping open sets forward from the domain to the codomain, meaning the image of every open set in $ X $ must be open in $ Y $. However, openness says nothing about the behavior of preimages in relation to open sets.

In summary: continuity deals with "pulling back" open sets, while openness deals with "pushing forward" open sets. They operate differently.

And so on.