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The Real And Complex Number Systems

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a in S and every r 0, the set x : dx, a r is nonempty.<br />

Proof: Assume that x : dx, a r is empty. Denote two sets x : dx, a r by A<br />

and x : dx, a r by B. <strong>The</strong>n we have<br />

1. A since a A and B since S is unbounded,<br />

2. A B ,<br />

3. A B S,<br />

4. A Ba; r is open in S,<br />

and consider B as follows. Since x : dx, a r is closed in S, B S x : dx, a r<br />

is open in S. So, we know that S is disconnected which is absurb. Hence, we know that the<br />

set x : dx, a r is nonempty.<br />

Supplement on a connected metric space<br />

Definition Two subsets A and B of a metric space X are said to be separated if both<br />

A clB and clA B .<br />

AsetE X is said to be connected if E is not a union of two nonempty separated<br />

sets.<br />

We now prove the definition of connected metric space is equivalent to this definiton<br />

as follows.<br />

<strong>The</strong>orem A set E in a metric space X is connected if, and only if E is not the union of<br />

two nonempty disjoint subsets, each of which is open in E.<br />

Proof: () Suppose that E is the union of two nonempty disjoint subsets, each<br />

of which is open in E, denote two sets, U and V. Claim that<br />

U clV clU V .<br />

Suppose NOT, i.e., x U clV. Thatis,thereisa 0 such that<br />

B X x, E B E x, U and B X x, V <br />

which implies that<br />

B X x, V B X x, V E<br />

B X x, E V<br />

U V ,<br />

a contradiction. So, we have U clV . Similarly for clU V . So,X is<br />

disconnected. That is, we have shown that if a set E in a metric space X is<br />

connected, then E is not the union of two nonempty disjoint subsets, each of which<br />

is open in E.<br />

() Suppose that E is disconnected, then E is a union of two nonempty<br />

separated sets, denoted E A B, whereA clB clA B . Claim that A<br />

and B are open in E. Suppose NOT, it means that there is a point x A which is<br />

not an interior point of A. So, for any ball B E x, r, there is a correspounding<br />

x r B, wherex r B E x, r. It implies that x clB which is absurb with<br />

A clB . So, we proved that A is open in E. Similarly, B is open in E. Hence,<br />

we have proved that if E is not the union of two nonempty disjoint subsets, each of<br />

whichisopeninE, then E in a metric space X is connected.<br />

Exercise Let A and B be connected sets in a metric space with A B not connected and<br />

suppose A B C 1 C 2 where clC 1 C 2 C 1 clC 2 . Show that<br />

B C 1 is connected.

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