Page 118 - DMTH503_TOPOLOGY

Topology

Notes Theorem 7: A subspace of the real line R is connected iff it is an interval. In particular, R is
connected.
Proof: Let E be a subspace of R.

We first prove that if E is connected, then it is an interval. Let us suppose that E is not an interval.
Then there exists real numbers a, b, c with a < c < b such that a, b  E but c  E.
Let A = ]–, C[ and B=]c, [.

Then A and B are open subset of R such that a  A and b  B.
Now, E A   and E B  , since a  E A and b  E B.
Also, (E A) (E B) = E (A B) =  ( A B = )
and (E A) (E B) = E (A B) = E R – {c} = E

Thus, A B forms a disconnection of E i.e., E is disconnected, a contradiction.
Hence E must be an interval.
Conversely, Let E be an interval and if possible let E is disconnected.
Then E is the union of two non-empty disjoint sets G and H, both closed in E, i.e. E = G H.

Let a  G and b  H
Since G H = , we have a  b
So either a < b or b < a
Without any loss of generality we may assume that a < b.

Since a, b  E and E is an interval, we have [a, b]  E = G H.
Let p = sup{G [a, b]}, then clearly a  p  b
Consequently, p  E.
But, G being closed in E, 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 E must be connected
Theorem 8: Prove that the real line is connected.

Proof: Let R be an interval and if possible let R is disconnected. Then R is the union of two
non-empty disjoint sets G and H, both closed in R, i.e. R = G H.
Let a  G and b  H.

Since G H = , we have a  b
So either a < b or b < a
Without any loss of generality, we may assume that a < b.
Since a, b  R and R is an interval, we have [a, b]  R = G H

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