Page 123 - DMTH503_TOPOLOGY

Unit 12: Components and Local Connectedness

Notes

Notes
(i) Let (X, J) be a topological space and E be a subset of X. If x  E, then the union of all
connected sets containing x and contained in E, is called component of E with respect
to x and is denoted by C (E, x).
(ii) Since the union of any family of connected sets having a non-empty intersection is
a connected set, therefore the component of E with respect of x i.e. C (E, x) is a
connected set.
(iii) If E is a component of X, the E .

Example 1:
(i) If X is a connected topological space, then X has only one component, namely X itself.
(ii) If X is a discrete topological space, then each singleton subset of X is its component.

Theorem 1: In a topological space (X, T) each point in X is contained in exactly one component
of X.
Proof: Let x be any point of X

Let A = {A } be the class of all connected subspaces of X which contains x
x i
A   as {x}  A
x x
Also (i) A   since x  A i
i
i
Therefore by theorem, Let X be a topological space and {A } be a non-empty class of connected
i
subspaces of X such that A   then A  U A i is connected subspace of X, A  C x (say)
i
i
i
i i
is connected subspace of X.
Further, x  C x and if B is any connected subspace of X containing x, then B  A x and so B  C .
x
Therefore C is a maximal connected subspace i.e. a component of X containing x.
x
Now we shall prove that C is the only component which contains x.
x
*

*
Let C be any other component of X which contain x. The C is one of the A s and is therefore
x
x
i
*
contained in C . But C is maximal as a connected sub-space of X, therefore we must have
x x
*
C  C i.e. C is unique in the sense that each point x  X is contained in exactly one component
x
x
x
C of X.
x
Theorem 2: In a topological space each components is closed.
Proof: Let (X, T) be a topological space and let C be a component of X.
By the definition of component, C is the largest connected set containing x. Then, C is also a
connected set containing x.
Thus C  C
Also C  C
Therefore C = C
Hence C is closed.

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