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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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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.
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.
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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