Math and science::Topology

Connectedness. 4 lemmas

1. The edge of a connected space

Let $X$ be a topological space. Let $A$ and $B$ be subspaces of $X$ with $A \subseteq B \subseteq Cl (A)$ .

If $A$ is connected, then so is $B$ .

2. An image of a continuous function on a connected space is connected

Let $f : X \to Y$ be a continuous map of topological spaces. If $X$ is connected then so is $f X$ .

In particular, any quotient of a connected space is connected.

3. The product of two connected spaces is connected.

4. A space that has an overlapping covering of connected subspaces is connected.

Let $X$ be a nonempty topological space and $(A_{i})_{i \in I}$ a family of subspaces covering $X$ . Suppose that $A_{i}$ is connected for each $i \in I$ and that $A_{i} \cap A_{j} \neq \emptyset$ for each $i, j \in I$ , then $X$ is connected.

This lemma says that gluing together overlapping connected spaces produces connected spaces.

1. ( $A is connected ⟹ A \subseteq B \subseteq C l (A)) ⟹ B is connected$ ). Proof outline.

$B$ is within the smallest closed subset containing $A$ . How could $B$ be divide into 2 disjoint closed/open sets with $A$ in one and the rest in another? It can't—if it could, then $A$ itself would be closed, and thus, it's own closure. Note: $A$ is connected, so cannot be subdivided to create the pair of open/closed subsets of $B$ .

A good example for visualizing this lemma is to consider $A$ to be the open unit disk, and for this lemma to represent the inability to create a separation of the closed disk with the open disk contained whollly in one of the sets of the separation.

2. A continuous image of a connected space is connected. Proof outline.

The restricted function $f : X \to f X$ is surjective, and $f^{- 1} f X = X$ . If we can split $f X$ into two disjoint non-empty open sets, then the pre-image of those sets will be open, disjoint and non-empty in $X$ , and they will union to $X$ . But that is a contradiction, as $X$ is connected.

3. Product of two connected spaces is connected. Proof outline.

Let $X$ and $Y$ be connected spaces, and consider $X \times Y$ .

Two approaches:

Show that if $X \times Y$ is not connected, then either $X$ or $Y$ cannot be connected. An ability to find a separation of $X \times Y$ implies that we have found a separation of either $X$ or $Y$ or both.
The approach taken by Leinster is to consider a continuous function $f : X \times Y \to D$ , where $D$ is a discrete space, and show that $f$ is a constant function. In the 'Connectedness. Equivalent formulations' card, it was shown that this is necessary and sufficient for $X \times Y$ to be connected. The demonstration that $f$ is constant is done by taking two arbitrary points $t_{1}, t_{2} \in X \times Y$ and considering a 'vertical' slice passing through $t_{1}$ and a 'horizontal' slice passing through $t_{2}$ . These slices are homeomorphic to one of $X$ and $Y$ , so $f$ must be constant alone both. As the two slices intersect somewhere, $f$ must be the same constant on each. As the points were arbitrary, this covers the whole space.

4. Union of overlapping connected spaces is connected. Proof.

Let $f$ be a continuous map from $X$ to a discrete space $D$ . For each $i \in I$ , we have the continuous map $f |_{A_{i}} : A_{i} \to D$ . $A_{i}$ is connected, so $f |_{A_{i}}$ has a constant value $d_{i}$ . But for each $i, j \in I$ , we have $A_{i} \cap A_{j} \neq \emptyset$ , so $d_{i} = d_{j}$ . Hence $d_{i}$ is independent of $i \in I$ , so $f$ is constant, as required.

Example: the letter O is a quotient of $[0, 1]$ , which is connected, so $O$ is connected. Gluing O to another line segment gives P, which must now also be connected. Gluing on another line segment to give A shows us that A too is connected.

Context

Source

Leinster, p70-71

Gowers does a mechanical step-by-step proof for pre-image of open is open through continuous function: https://www.dpmms.cam.ac.uk/~wtg10/easyanalysis1.html