The converse is not the case, so a sequence may be Cauchy without being convergent.
Consider the spacewith the absolute value metric. The sequence is defined as follows:
Each termin the sequence toto n significant figures. This sequence converges to butsodoes not converge.
]]>Ifare Cauchy sequences in a metric spacesuch thatforthen
is also a Cauchy sequence
converges toif and only ifconverges to
Proof
For 1:
Applying the triangle inequality,
LetWe can findsuch that
Sinceis Cauchy, there existssuch that
Now letwe gethence is Cauchy.
For 2:
Using the triangle theorem again gives
Hence,but
Ifthen and
Similarly, ifthen
]]>Letbe the positive integers with topology consisting of the open setsand letbe a non empty subset of
LetIfis even thenis a limit point ofand ifis odd thenis a limit point of
Hencehas an accumulation point andis countably compact.
The setsconstitute an open cover ofand is not reducible to a finite subcover. Henceis not compact.
]]>Suppose we have a metric spaceSupposeis a Cauchy sequence in
Defineas follows:where
is dense in the quotient spacewhereis the equivalence relation on the set of Cauchy sequences indefined by:
if
thenis dense in
Proof
Let
Thenis a Cauchy sequence in
Henceis the limit of the sequenceinHenceis dense in
]]>The family of open intervalshas the finite intersection property.
Proof
A familyof sets is said to have the finite intersection property if every finite collectionhas a non empty intersection, so that
Letbe a subset of
ifsoThis can obviously be extended inductively to any finite intersection.
Letthen
]]>Now supposeis a Cauchy sequence in the space. LetThere exists such thatso that
Hence a Cauchy sequence in this space takes the form
]]>To show this, consider
Takeand
]]>Consider the setwith topology
The closed sets areand
The topological spaceis a regular space butis not a T1 space since the singleton setis not closed:
]]>Consider the set the graph of which is shown below.
For anyconsists of an infinite disconnected sections of curve sois connected but not locally connected.
is a continuous image of the connected setthe property of being locally connected is not a topological property.
]]>Define the topologyon
Sets containing 1 are open if and only if they are the complement of a finite set.
Sets not containing 1 are open when they are open in setr with the usual topology.
Hence, each open set containing 0 is infinite.
Therefore points 0 and 1 cannot be placed in open disjoint sets.
The space is T1 but not T2.
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