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Unit 11: Connected Spaces, Connected Subspaces of Real Line
Now from (1) and (2), we get Notes
H = ,
which contradicts the given fact that H is non-empty.
Hence A is also a connected set.
Theorem 3: If every two points of a set E are contained in same connected subset of E, then E is
connected.
Proof: Let us suppose that E is not connected.
Then, it must a separation E = A|B
i.e. E is the union of non-empty separated sets A and B.
Since A and B are non-empty, let a A and b B.
Then, A and B being disjoint
a, b are two distinct points of E.
So, by given hypothesis there exists a connected subset C of E such that a, b c
But, C being a connected subset of a disconnected set E with the separation E = A|B,
we have C A or C B.
This is not possible, since A and B are disjoint and C contains at least one point of A and one that
of B, which leads to a contradiction.
Hence E is connected.
Theorem 4: A topological space (X, T) is connected iff the only non-empty subset of X which is
open and closed is X itself.
Proof: Let (X, T) be a connected space.
c
Let A be a non-empty subset of X that is both open and closed. Then A is both open and closed.
C
A = A and A = A C
C
Thus A A =
C
C
A A = and A A =
Also X = A A C
C
Therefore A and A are two separated sets whose union in X.
C
C
Now if A and A , then we have separation X = A|A , which leads to the contradiction as
X is connected.
So either A = or A =
C
C
But A = or A =
But A
So A =
C
X = A A = A = A
C
This shows that the only non-empty subset of X that is both open and closed is X itself. Conversely,
let the only subset of X which is both open and closed be X itself.
Then, there exists no non-empty proper subset of X which is both open and closed.
Hence (X, T) is not disconnected and therefore, it is connected.
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