Preservation of Connectedness under Continuous Maps

If $ X $ is a connected topological space and $ f : X \to Y $ is a continuous function, then $ f(X) $ is a connected subset of $ Y $.

Equivalently, the continuous image of a connected set remains connected.

If we begin with a connected space $ X $ and apply a continuous function $ f $, the resulting set of points $ f(X) $ cannot lose its connectedness. Continuity alone is not sufficient to break a space into separate pieces.

In this precise sense, connectedness is preserved.

What does connected mean? A topological space is said to be connected if it cannot be written as the union of two nonempty, disjoint open sets. For instance, a line segment is a connected set of points. By contrast, two isolated points form a disconnected space.

A concrete example
The proof

A concrete example

Consider the topological space

$$ X = [0,1] \subset \mathbb{R} $$

The closed interval $ [0,1] $ is connected. Intuitively, it forms a single continuous piece, with no gaps or separations.

Now define a function $ f : [0,1] \rightarrow \mathbb{R} $

$$ f(x) = 2x $$

This function is continuous, and the image of the interval is

$$ f([0,1]) = [0,2] $$

The image $ f(X) = [0,2] $ is also connected.

Thus, connectedness is preserved.

Note. To show that a set is not connected, one would have to find two open sets $ U $ and $ V $, disjoint ( $ U \cap V = \emptyset $ ) and nonempty ( $ U \ne \emptyset $, $ V \ne \emptyset $ ), whose union covers the entire set $ f(X) $, that is $ f(X) \subset U \cup V $. This cannot occur here, because any pair of disjoint open sets intersecting $ [0,2] $ would necessarily exclude at least one point of the interval $ [0,2] $. Such a situation would amount to an impossible separation of a real interval. Therefore, $ [0,2] $ is connected.

Example 2

Consider again the topological space $ X $

$$ X = [0,1] \subset \mathbb{R} $$

The interval $ [0,1] $ is connected.

Define the function $ f : [0,1] \rightarrow \mathbb{R} $

$$ f(x) = 0 $$

The function $ f $ is continuous, and its image is

$$ f(X) = \{ 0 \} $$

Geometrically, the interval $ [0,1] $ is collapsed, contracting entirely to a single point ($ 0 $).

Nevertheless, the image $ f(X) $ remains connected, since the set $ \{ 0 \} $ is nonempty, consists of a single element, and cannot be decomposed into two disjoint subsets.

Accordingly, connectedness of the image is preserved.

Note. The function has folded the interval, but it has not broken it. Any continuous function defined on an interval cannot split it into separate pieces, nor can it produce a disconnected set. It may compress the space, identify distinct points, or reduce its dimension, but it cannot destroy connectedness. Producing a disconnected image would require a discontinuity.

The proof

The argument proceeds by contradiction.

Assume that $ X $ is a connected space, but that its continuous image $ f(X) $ is not connected.

If $ f(X) $ is not connected, then there exist two open sets $ U $ and $ V $ that form a separation of $ f(X) $. That is, $ f(X) \subset U \cup V $, and every point of $ f(X) $ belongs either to $ U $ or to $ V $, but not to both.

This is the crucial observation. Since $ f $ is continuous, and the preimage of an open set under a continuous function is open, it follows that:

$ f^{-1}(U) $ is open in $ X $
$ f^{-1}(V) $ is open in $ X $

A contradiction now emerges.

Because $ U $ and $ V $ contain points of $ f(X) $, the sets $ f^{-1}(U) $ and $ f^{-1}(V) $ are nonempty, disjoint, and together cover the entire space $ X $.

This implies that $ X $ can be written as the union of two nonempty, disjoint open sets.

But this contradicts the assumption that $ X $ is connected.

The contradiction shows that the initial assumption is false. Consequently, the opposite statement holds: the continuous image of a connected set is connected.

Note. Put differently, a continuous function may bend or compress a space, but it cannot create gaps or tears. Splitting a space into separate pieces requires a discontinuity.

And so on.