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Connectivity
Exercise 5.1. Let A = (Q ⇥ R) [ (R ⇥ Q) ⇢ E2 be the subset of points with at least one
rational coordinate. Show that A is path-connected, hence connected.
Recall that a space A is path-connected if there is a path : [0, 1] ! A between any two
points. Since paths can be concatenated it is enough to show that there is a path from (0, 0)
to each point, for then we can go from (a, b) to (c, d) by going via (0, 0).
Suppose (p, y) 2 Q ⇥ R. Then there is a path t 7! (p, ty) in Q ⇥ R from (p, 0) to (p, y)
and also a path t 7! (tp, 0) in R ⇥ Q from (0, 0) to (p, 0). Concatenating these gives a path
from (0, 0) to (p, y) in the union (Q⇥ R) [ (R⇥Q) = A.
The other case for a point (x, q) 2 R ⇥ Q is similar but with the coordinates swapped.
Hence A is path-connected, and therefore also connected.
Exercise 5.2. Find the components of R in the finite-complement topology.
Let R have the finite-complement topology in which U ⇢ R is open () U = ; or the
complement X U is a finite set. I claim that R is connected in this topology and so has one
connected component, namely R itself. To see that this is so, suppose R = U [ V where U
and V are open and non-empty. Since U, V 6= ; both RU and R V are finite sets. Hence
U \ V is non-empty because
R (U \ V ) = (R U) [ (R V )
is the union of two finite sets, hence is itself finite, whereas R is an infinite set. This shows
that R is connected in the finite-complement topology.
[The same proof shows that if X has infinitely points then X is connected in the finite-
complement topology. In contrast, ifX has only finitely many points then the finite-complement
topology is the discrete one in which every subset is open, so X is disconnected if it has two
or more points. Of course, any topology on the empty set or a single point is connected.]
Exercise 5.3. Show that Q with the Euclidean topology is totally-disconnected.
Suppose that A ⇢ Q is connected in the Euclidean topology. Since the inclusion Q ,! E
is continuous, and the continuous image of a connected set is connected, we deduce that A is
also connected as a subset of E. We have seen that this means A is an interval. As A contains
no irrational points the only possibility is that A = [q, q] = {q} for some q 2 Q. Therefore the
only connected subsets of Q in the Euclidean topology are the singletons, and this is exactly
what it means for Q to be totally disconnected.
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Exercise 5.4. Suppose A ⇢ X is a dense, connected subset. Show that X is connected. Deduce
that the closure of a connected subspace is connected.
Suppose X = U [V where U and V are open and non-empty. Then the intersection A\U
of U with the dense subset A is non-empty, for otherwise the closure A ⇢ X U is not the
whole of X. Likewise A \ V 6= ;. So A = (A \ U) [ (A \ V ) where A \ U and A \ V are
open (in A) and non-empty. Since A is connected, the intersection (A \ U) \ (A \ V ) 6= ;,
and hence U \ V 6= ; either. This shows that X cannot be disconnected, i.e. X is connected.
By definition if B ⇢ X then B is dense in its closure B. Therefore if B is connected so is
its closure.
Exercise 5.5. Suppose A and B are path-connected and A \ B is non-empty. Deduce that
A [ B is path-connected.
This follows because the concatenation of two paths is again a path. In more detail: fix a
point x 2 A \ B. If y 2 A [ B then either y 2 A which is path-connected, so there is a path
from y to x in A, or y 2 B which is also path-connected, so there is a path from y to x in B.
If z is another point in A[B then there is likewise a path from x to z. Concatenating the path
from y to x with that from x to z gives a path from y to z. Hence A [ B is path-connected
as claimed.
Exercise 5.6. Give an example of a path-connected subspace whose closure is not path-connected.
The subspace X = {(x, sin(1/x)) | x > 0} ⇢ E2 is path-connected, because it is home-
omorphic to (0,1) ⇢ E via the projection (x, y) 7! x. It’s closure is the topologists’ sine
curve
X = {(0, y) | 1 y 1} [ {(x, sin(1/x)) | x > 0} ⇢ E2
which is connected but not path-connected.