Show that a set A in ℝ2 is open in the Euclidean metric ⇔ it is open in the max metric.
Show that a set A in ℝ2 is open in the Euclidean metric ⇔ it is open in the max metric.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Show that a set A in ℝ2 is open in the Euclidean metric ⇔ it is open in the max metric.
Hint: As usual, there are two directions to prove in an ⇔. The picture on p73 of the notes may be somewhat helpful.
![Cantionary tales Although sn= (-11"
Indeed, since
this is
Examples
Example
Definitions We say
while another
This situation
with
We introduce terminology!
that
is
has convergent subsequences
subsequenos
Conso
Converse to -1,
another prout that
The
the
One
metriz
(-)"
(-1)".n
Definition If a subsequence of
dive-ses
to
real
and
i) has an
40
ii)
has
,A
(-11" is important and interesting,
a
space
r
as an
accumulation print if r=lim Snk for some
subsequence (S) of (su).
say
Lemma A recl-valmed.
not convergent, it
ZK
Sequence
(M,d) has
A
(sn) has
(-1)
(-1)" is dit-gent,
and (-1) 2KH
to 1 Las a constant sey)
(su) of points in
1
an
Lor an
real-valued sequence (su)
(or to-co),
accumulation points at 1,-1
AL
44
76
accumulation point at co
"
40
at co, -00,
put of
i^
is the
sequeare
has accumulation points at 1,0-1,
4
(1, 0,-1,0,1,...)
Accumulation points at I wo give us an alternative way
to talk about unbunded sequences, as the following lemma tells us.
(S₂)
sequence
accumulatim punt at a
.).
-Co
(sa) has no bound from above, and
(sa) has no bund from below.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5be4dcbe-6a78-40bc-9e54-3592389d165a%2F2eb53b10-2ff2-461b-a667-d90ab1c3b5a9%2F4ev8oocb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Cantionary tales Although sn= (-11"
Indeed, since
this is
Examples
Example
Definitions We say
while another
This situation
with
We introduce terminology!
that
is
has convergent subsequences
subsequenos
Conso
Converse to -1,
another prout that
The
the
One
metriz
(-)"
(-1)".n
Definition If a subsequence of
dive-ses
to
real
and
i) has an
40
ii)
has
,A
(-11" is important and interesting,
a
space
r
as an
accumulation print if r=lim Snk for some
subsequence (S) of (su).
say
Lemma A recl-valmed.
not convergent, it
ZK
Sequence
(M,d) has
A
(sn) has
(-1)
(-1)" is dit-gent,
and (-1) 2KH
to 1 Las a constant sey)
(su) of points in
1
an
Lor an
real-valued sequence (su)
(or to-co),
accumulation points at 1,-1
AL
44
76
accumulation point at co
"
40
at co, -00,
put of
i^
is the
sequeare
has accumulation points at 1,0-1,
4
(1, 0,-1,0,1,...)
Accumulation points at I wo give us an alternative way
to talk about unbunded sequences, as the following lemma tells us.
(S₂)
sequence
accumulatim punt at a
.).
-Co
(sa) has no bound from above, and
(sa) has no bund from below.
![Profi First, notice
k
and
So it suffices to pone (i). We verify (=) and (5).
As a
subsequence
Now
It follows that
دی
77
that (ii) fullows from (i) by multiplying by -1;
=co) lim-S₁ ===
So
Sux
VM kw >> KW/S₁ <-M.
natural
(The set of such
E): Recursively define the increasing subsequence
by n₁ = 1, and so
NICH
am
(as 3e wal
Smok
IN
Sme diveger to co
[K>N] =>
{nx >M.
arbitrary M is not in bond from above for for
Sme>M.)
(strate))
with neers > Moe and SMK
k→
numbus
is well-defird
Suk
hr of IN
that he is the least
is
and unbranded as
= co
is nunempty since
increasing by construction
"
a)
So
max(KH, Sn₂)
Sn is unlunded,
IN is well-ordered).
Or, just take
N=Min limit
definition!
Sny > Me
Sn has accumulation pt at co.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5be4dcbe-6a78-40bc-9e54-3592389d165a%2F2eb53b10-2ff2-461b-a667-d90ab1c3b5a9%2F8p021uo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Profi First, notice
k
and
So it suffices to pone (i). We verify (=) and (5).
As a
subsequence
Now
It follows that
دی
77
that (ii) fullows from (i) by multiplying by -1;
=co) lim-S₁ ===
So
Sux
VM kw >> KW/S₁ <-M.
natural
(The set of such
E): Recursively define the increasing subsequence
by n₁ = 1, and so
NICH
am
(as 3e wal
Smok
IN
Sme diveger to co
[K>N] =>
{nx >M.
arbitrary M is not in bond from above for for
Sme>M.)
(strate))
with neers > Moe and SMK
k→
numbus
is well-defird
Suk
hr of IN
that he is the least
is
and unbranded as
= co
is nunempty since
increasing by construction
"
a)
So
max(KH, Sn₂)
Sn is unlunded,
IN is well-ordered).
Or, just take
N=Min limit
definition!
Sny > Me
Sn has accumulation pt at co.
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