Page 127 - DMTH503_TOPOLOGY

Unit 12: Components and Local Connectedness

But A B is not locally connected at (0, 1), since the open disc with centre (0, 1) and radius Notes
 1  2
  does not contain any connected open subset of R containing (0, 1).
 4 
Hence A B is connected but not locally connected.
Theorem 8: Every component of a locally connected space is open.
Proof: Let (X, T) be a locally connected space and E be a component of X.
We shall show that E is an open set.

Let x be any element of E.
Since X is locally connected, there exists a connected space set G which contains x. Since E is a
x
component, we have x G E clearly, E = {G : x  E}.
x x
Therefore E, being a union of open sets, is an open set.
Theorem 9: A topological space X is locally connected iff the components of every open subspace
of X are open in X.

Proof: Let X be locally connected and Y be an open subspace of X.
Let E be a component of Y.
We are to show that E is open in X i.e. if x is any element of E then there exists a nhd. G of x such
that G E.
Now E , Y, Y open in X, x Y and X is locally connected implies that there exists a connected
open set G containing x such that G Y.
Since the topology which G has as a subspace of Y is the same as that it has as a subspace of X,
therefore G is also connected as a subspace of Y and consequently G E as E is a component of Y.
Conversely, let the components of every open subspace of X be open in X, Let x X and Y an
open subset of X containing x. Let E be a component of Y containing x. Then by hypothesis, E
x x
is open and connected in Y and therefore in X.

Example 5: Give an example of locally connected space which is totally disconnected.
Solution: Every discrete space is locally connected as well as totally disconnected.

Let x be an arbitrary point of a discrete space X.
We know that every subset of a discrete space is open and that every singleton set is connected.
Hence {x} is a connected open nhd. of x. Also every open nhd. of x must contain {x}.

Hence X is locally connected.
To prove X is totally disconnected.
Let x, y be any two distinct points of a discrete space X.
The G = {x} and H = X – {x} are both non-empty open disjoint sets whose union is X such that x  G
and y  H. It follows that X is totally disconnected.
Theorem 10: Local connectedness neither implies nor is implied by connectedness.
Proof: The union of two disjoint open intervals on the real line forms a space which is locally
connected but not connected. Example of a space which is connected but not locally connected.

Let X be the subspace of Euclidean plane defined by
X = A  B where

LOVELY PROFESSIONAL UNIVERSITY 121