Lemma 3.3.3. If X is a normed space and xn, x, Yn, y are vectors in X, then the following statements hold. (a) Uniqueness of limits: If xn → x and xn → y, then x = y. (b) Reverse Triangle Inequality: ||||| – ||y||| < ||x – y|l. (c) Convergent implies Cauchy: If xn → x, then {xn}n€N is Cauchy.

Algebra & Trigonometry with Analytic Geometry
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Chapter9: Systems Of Equations And Inequalities
Section9.9: Properties Of Determinants
Problem 46E
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Lemma 3.3.3. If X is a normed space and xn, x, Yn, y are vectors in X,
then the following statements hold.
(a) Uniqueness of limits: If xn
→ x and xn - y, then x = y.
(b) Reverse Triangle Inequality: |||x|| – |||| < ||x – y||
(c) Convergent implies Cauchy: If xn → x, then {xn}nɛN is Cauchy.
(d) Boundedness of Cauchy sequences: If {xn}nɛN is a Cauchy sequence, then
sup ||xn|| < o.
(e) Continuity of the norm: If xn
(f) Continuity of vector addition: If x,
→ x, then ||xn || → ||r||.
→ x and yn
y, then xn+yn → x+y.
(g) Continuity of scalar multiplication: If xn → x and cn → c (where cn, c
are scalars), then cnan → cx.
Proof. We will prove one part, and assign the proof of the remaining state-
ments as Problem 3.3.8.
(d) Suppose that {xn}nɛN is Cauchy. Then, by applying the definition of
a Cauchy sequence with the specific choice e = 1, we see that there exists
some N > 0 such that ||xm – xn|| < 1 for all m, n > N. Therefore, for all
n > N we have
||rn||
||Ln – IN +IN|| < ||xn – ¤ || + ||æ|| < 1 + ||®N||-
%3D
3.3 The Induced Metric
113
Hence, for an arbitrary n we have
||Tn|| < max{||x1||, ... |"N-1||, ||TN|| +1}.
Thus {xn}neN is a bounded sequence.
Since a normed space is a vector space, we can define lines, planes, and
other related notions. In particular, if x and y are vectors in X, then the line
segment joining x to y is the set of all points of the form tx + (1 – t)y, where
0 <t< 1.
Transcribed Image Text:Lemma 3.3.3. If X is a normed space and xn, x, Yn, y are vectors in X, then the following statements hold. (a) Uniqueness of limits: If xn → x and xn - y, then x = y. (b) Reverse Triangle Inequality: |||x|| – |||| < ||x – y|| (c) Convergent implies Cauchy: If xn → x, then {xn}nɛN is Cauchy. (d) Boundedness of Cauchy sequences: If {xn}nɛN is a Cauchy sequence, then sup ||xn|| < o. (e) Continuity of the norm: If xn (f) Continuity of vector addition: If x, → x, then ||xn || → ||r||. → x and yn y, then xn+yn → x+y. (g) Continuity of scalar multiplication: If xn → x and cn → c (where cn, c are scalars), then cnan → cx. Proof. We will prove one part, and assign the proof of the remaining state- ments as Problem 3.3.8. (d) Suppose that {xn}nɛN is Cauchy. Then, by applying the definition of a Cauchy sequence with the specific choice e = 1, we see that there exists some N > 0 such that ||xm – xn|| < 1 for all m, n > N. Therefore, for all n > N we have ||rn|| ||Ln – IN +IN|| < ||xn – ¤ || + ||æ|| < 1 + ||®N||- %3D 3.3 The Induced Metric 113 Hence, for an arbitrary n we have ||Tn|| < max{||x1||, ... |"N-1||, ||TN|| +1}. Thus {xn}neN is a bounded sequence. Since a normed space is a vector space, we can define lines, planes, and other related notions. In particular, if x and y are vectors in X, then the line segment joining x to y is the set of all points of the form tx + (1 – t)y, where 0 <t< 1.
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