Page 119 - DMTH503_TOPOLOGY

Unit 11: Connected Spaces, Connected Subspaces of Real Line

Let p = sup{G [a, b]}, then clearly a  p  b Notes
Consequently p  R
But G being closed in R, the definition of p shows that p  G and therefore p  b.

Consequently, p < b
Moreover the definition of p shows that p +   H for each  > 0 for which p +   b.

This shows that every nhd. of p contains at least one point of H other than p. So p is a limit point
of H.
But H being closed, we have p  H

Thus p  G H and therefore G H  , which is a contradiction.
Hence R must be connected.

Example 4: Show that if X is a connected topological space and f is a non-constant
continuous real function defined on X then X is uncountably infinite.

Solution: f : X   is continuous and X is connected, so f (X) is a connected subspace of .
Suppose that f (X) is not connected, there exists a non-empty proper subset E of f (X) such that E
is both open and closed in f (X).
As f is continuous
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 f (E) is non-empty proper subset of X which is both open and closed in X.
This contradicts the fact that X is connected. Hence f (X) must be a connected subspace of .
Also f is non-constant, there exist x, y  X such that f (x)  f (y)

Let a = f (x) and b = f (y).
Without any loss of generality we may suppose that a < b. Now a, b  f (X), f (x) is a connected
subspace of 

 [a, b]  f (X).
[ a subspace E of real line  is connected iff E is an interval
i.e. if a, b  E and a < c < b then c  E. In particular  is connected.]
Since [a, b] is uncountably infinite, it follows that f (X) is uncountably infinite and consequently
X must be uncountably infinite.

Example 5: Show that the graph of a continuous real function defined on an interval is a
connected subspace of the Euclidean plane.
Solution: Let f : 1   be continuous and let G be the graph of f.
Then G = I × f (I)   .
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Now since I is connected by the theorem “A subspace E of the real line  is connected iff E is an
interval.”

Also, f is continuous, it follows that f (I) is a connected subspace of R since continuous image of
a connected space is connected. Also we know that connectedness is a product invariant property,
hence G is connected.

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