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1. True or false? (Give reasons!)
(a) A closed subspace of a connected space is connected.
(b) The inverse image of a connected set under a continuous function is connected.
(c)A topological space with one point is connected.
(d) A topological space with two points is disconnected.
(e) R2 is not homeomorphic to R.
(a) False: e.g., [0, 1] ∪ [2, 3] in R. (Other counterexamples include any connected Hausdorff
space with two distinct points.)
(b) False: f : R → R given by x 7→ x2 is continuous, but {1} is connected, while f ∗ ({1}) =
{−1, 1} has the discrete topology and so is disconnected. (Others include non-injective
maps with discrete domain.)
(c) True: any continuous function on a one point space is constant.
(d) False: if X = {x, y} with the trivial topology, then the only open sets are X and ∅
and so there is no proper open partition.
(e) True: suppose for a contradiction that f : R → R2 is a homeomorphism. The restriction
to R \ {0} is a homeomorphism f : R \ {0} → R2 \ f (0). However, R \ {0} is disconnected,
whereas R2 \ f (0) is path-connected and hence connected.
2. Let Sα (α ∈ A) be subspaces of a topological space X such that α∈A Sα 6= ∅ and let
T = α∈A Sα . [Hint: it helps to note that if x, y ∈ T then there exist β, γ ∈ A and t ∈ T
with x ∈ Sβ , y ∈ Sγ and z ∈ Sβ ∩ Sγ .] Prove the following.
(a) If Sα is connected for all α ∈ A, then T is connected.
(b) If Sα is path-connected for all α ∈ A, then T is path-connected. [Hint: patch continuous
functions on a cover of a space by closed sets.]
(a) If Sα are connected, then any continuous function f : T → D to a discrete space must be
constant on Sα for all α ∈ A. In particular if x, y ∈ T and t is as above, f (x) = f (t) = f (y).
Thus any such f is constant on T . Thus T is connected.
(b) If Sα are path-connected and x, y ∈ T , then we can choose continuous paths h: [0, 1] →
Sβ and g: [0, 1] → Sγ with h(0) = t, h(1) = y, g(0) = x and g(1) = t. Then we form the
concatenation h#g: [0, 1] → T of the two paths by defining
for t ∈ [0, 1/2]
(h#g)(s) =
g(2s − 1) for t ∈ [1/2, 1]
Then h#g is a continuous path from x to y, by the patching for finite closed covers. Thus
T is path-connected.
3. Prove the following statements:
(a) the product of two path-connected spaces is path-connected.
(b) the product of two totally separated spaces is totally separated.
(c) the continuous image of a path-connected space is path-connected.
(d) the continuous image of a totally separated space may not be totally separated; indeed,
that it may be connected.
(e) a totally separated space is Hausdorff.
(a) Consider any two points (x1 , y1 ) and (x2 , y2 ) in X × Y . Since X is path-connected,
there is a continuous function f : [0, 1] → X with f (0) = x1 , f (1) = x2 . Since Y is pathconnected, there is a continuous function g: [0, 1] → Y with g(0) = y1 , g(1) = y2 . Then the
function h: [0, 1] → X × Y with h(s) = (f (s), g(s)) is continuous (since the composition
with either projection map is continuous) and satisfies h(0) = (x1 , y1 ), h(1) = (x2 , y2 ).
Thus X × Y is path-connected.
(b) Suppose that (x1 , y1 ) and (x2 , y2 ) are two distinct points in X × Y . Then wlog x1 6=
x2 . In the first case, since X is totally separated, there is a continuous f : X → D with
f (x1 ) 6= f (x2 ). Hence F = f ◦ π1 : X × Y → D is continuous with F (x1 , y1 ) = f (x1 ) 6=
f (x2 ) = F (x2 , y2 ).
(c) Suppose X is path-connected and f : X → Y is continuous. For any two points y1 , y2
in f (X), choose corresponding points x1 , x2 in X such that yi = f (xi ). Then, as X is
path-connected, there is a continuous path h: [0, 1] → X from x1 to x2 and therefore f ◦ h
is a continuous path from y1 to y2 . Thus f (X) is path-connected.
(d) A constant map has connected image, but unfortunately a one point space is also
totally separated for vacuous reasons (it is the only space which is both connected and
totally separated). Another very general example is any non-constant map from a totally
separated space onto a space with the trivial topology. This is automatically continuous
and has image which is connected but not totally separated.
(e) If x 6= y in a totally separated space X, then there is an open partition X = U ∪ V
with x ∈ U and y ∈ V . Keeping the condition that U and V are disjoint, but dropping
the requirement that U and V cover X, we obtain precisely the definition of Hausdorff.
4. Consider the subspace X of R2 which is the union of the sets {(1/n, y) | y ∈ [−1, 1]}, for
n ∈ Z+ and the two points p = (0, 1) and q = (0, −1). Show that p and q are in different
connected components of X, but that any continuous function f : X → D, where D is a
discrete space, must take the same value at p and q.
Solution: In fact, we can show that the connected components of X are {p} and {q}
and Yn = {(1/n, y) : y ∈ [−1, 1]} for each n. Observe first that each Yn is connected and
both open and closed in the subspace topology. Hence each Yn is a connected component
and therefore no point of Yn can be in the connected component of p. Furthermore,
the connected component of p can not be {p, q} which is a discrete subspace and hence
disconnected. Thus the connected component of p must be {p}, and similarly the connected
component of q must be {q}. (Note that {p} is closed, but not open, in the subspace
topology, so is not an ‘isolated’ component as Yn is.)
On the other hand, let f : X → D be a continuous map to a discrete space. Then f must
be constant on each Yn and in particular f (1/n, 1) = f (1/n, −1). Now, as n → ∞, we
have (1/n, 1) → p and (1/n, −1) → q and so f (1/n, 1) → f (p) and f (1/n, −1) → f (q), by
continuity of f . But a discrete space is Hausdorff, so f (p) = f (q) by the uniqueness of
GKS, 30/3/17