Page 116 - DMTH503_TOPOLOGY

Topology

Notes Theorem 5: A continuous image of connected space is connected.
Proof: Let f : X  Y be a continuous mapping of a connected space X into an arbitrary topological
space Y.
We shall show that f[X] is connected as a subspace of Y.

Let us suppose f[X] is disconnected.
Then there exists G and H both open in Y such that

G f[X]  , H f[X]  
(G f[X]) (H f[X]) = 
and (G f[X]) (H f[X]) = f[X]

It follows that
 = f []
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= f [(G f[X]) (H f[X])]
= f ((G H) f[X])
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= f [G] f [H] f (f[X])
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= f [G] f [H] X
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= f [G] f [H]
and X = f (f[X])
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= f [(G f[X]) (H f[X])]
= f [(G H) f[X])]
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= f [G H] f (f[X])
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= f [G] f [H] X
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= f [G] f [H]
Since f is continuous and G and H are open in Y both intersecting f[X].
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If follows that f [G] and f [H] are both non-empty open subsets of X.
Thus X has been expressed as union of two disjoint open subsets of X and consequently X is
disconnected, which is a contradiction.
Hence f[X] must be connected.

Example 2: Show that (X, T) is connected space if X = {a, b, c, d} and T = {X, , {a, b}}.
Solution: T-open sets are X, , {a, b}.
T-closed sets are , X, {c, d}

For X-{a, b} = {c, d}
Thus  non-proper subset of X which is both open and closed. Consequently (X, T) is not
disconnected. It follows that (X, T) is connected.

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