Topology of $(-2,-1) cup D[0,1] cup 1,2$

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.29,30

---

Definitions:

-

---

If I understood the either, or, and, etc right, then $G=(-2,-1) cup D[0,1] cup 1,2$ and thus

• 1.29 (a) $G$ is from left to right, an interval in $mathbb R$, the unit disc, a point in $mathbb R$ and then another point in $mathbb R$.


• 1.29 (b) Interior: $D[0,1]$


• 1.29 (c) Boundary: $[-2,-1] cup C[0,1] cup 1,2 = [-2,-1) cup C[0,1] cup 2$


• 1.29 (d) Isolated: $2$

1.30

Let $A=(-2,-1), B=D[0,1],C=1,D=2$. I'm gonna group these into $G_1$ and $G_2$ separated s.t. $G_1 subseteq G_3$ and $G_2 subseteq G_4$

Ways:

1. $G_1=A cup B cup C, G_2 = D, G_3 = x < 1.5, G_4 = x > 1.5$


2. $G_1=A cup D, G_2 = B cup C, G_3 = x < -1 cup x > 1.5, G_4 = -1 < x < 1.5$


3. $G_1=A, G_2 = B cup C cup D, G_3 = x < -1, G_4 = x > -1$

---

Where have I gone wrong please?

general-topology complex-analysis

share|cite|improve this question

edited Aug 5 at 11:51

asked Jul 29 at 11:52

BCLC

6,95921973

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did they really write $-2<z<-1$? I'd get another book....
  – Lord Shark the Unknown
  Jul 29 at 11:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @LordSharktheUnknown Why? It's a perfectly valid property of $z in mathbbC$ to say that $z$ is real and $-2 < z < -1$.
  – Adayah
  Jul 29 at 11:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Two places. First, note a boundary point of G need not be a member of G. Look again at -2. Second, none of your separations split B and C. That's correct, but suggests you should look again at the isolation of 1.
  – David Hartley
  Jul 29 at 12:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidHartley Thanks! ^-^ I actually noticed those transcribed incorrectly. Post as answer?
  – BCLC
  Jul 29 at 12:17
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What does the notation $D[0,1]$ mean?
  – Henning Makholm
  Jul 29 at 13:40
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

favorite

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.29,30

---

Definitions:

-

---

If I understood the either, or, and, etc right, then $G=(-2,-1) cup D[0,1] cup 1,2$ and thus

• 1.29 (a) $G$ is from left to right, an interval in $mathbb R$, the unit disc, a point in $mathbb R$ and then another point in $mathbb R$.


• 1.29 (b) Interior: $D[0,1]$


• 1.29 (c) Boundary: $[-2,-1] cup C[0,1] cup 1,2 = [-2,-1) cup C[0,1] cup 2$


• 1.29 (d) Isolated: $2$

1.30

Let $A=(-2,-1), B=D[0,1],C=1,D=2$. I'm gonna group these into $G_1$ and $G_2$ separated s.t. $G_1 subseteq G_3$ and $G_2 subseteq G_4$

Ways:

1. $G_1=A cup B cup C, G_2 = D, G_3 = x < 1.5, G_4 = x > 1.5$


2. $G_1=A cup D, G_2 = B cup C, G_3 = x < -1 cup x > 1.5, G_4 = -1 < x < 1.5$


3. $G_1=A, G_2 = B cup C cup D, G_3 = x < -1, G_4 = x > -1$

---

Where have I gone wrong please?

general-topology complex-analysis

share|cite|improve this question

edited Aug 5 at 11:51

asked Jul 29 at 11:52

BCLC

6,95921973

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did they really write $-2<z<-1$? I'd get another book....
  – Lord Shark the Unknown
  Jul 29 at 11:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @LordSharktheUnknown Why? It's a perfectly valid property of $z in mathbbC$ to say that $z$ is real and $-2 < z < -1$.
  – Adayah
  Jul 29 at 11:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Two places. First, note a boundary point of G need not be a member of G. Look again at -2. Second, none of your separations split B and C. That's correct, but suggests you should look again at the isolation of 1.
  – David Hartley
  Jul 29 at 12:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidHartley Thanks! ^-^ I actually noticed those transcribed incorrectly. Post as answer?
  – BCLC
  Jul 29 at 12:17
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What does the notation $D[0,1]$ mean?
  – Henning Makholm
  Jul 29 at 13:40
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

favorite

up vote
1
down vote

favorite

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.29,30

---

Definitions:

-

---

If I understood the either, or, and, etc right, then $G=(-2,-1) cup D[0,1] cup 1,2$ and thus

• 1.29 (a) $G$ is from left to right, an interval in $mathbb R$, the unit disc, a point in $mathbb R$ and then another point in $mathbb R$.


• 1.29 (b) Interior: $D[0,1]$


• 1.29 (c) Boundary: $[-2,-1] cup C[0,1] cup 1,2 = [-2,-1) cup C[0,1] cup 2$


• 1.29 (d) Isolated: $2$

1.30

Let $A=(-2,-1), B=D[0,1],C=1,D=2$. I'm gonna group these into $G_1$ and $G_2$ separated s.t. $G_1 subseteq G_3$ and $G_2 subseteq G_4$

Ways:

1. $G_1=A cup B cup C, G_2 = D, G_3 = x < 1.5, G_4 = x > 1.5$


2. $G_1=A cup D, G_2 = B cup C, G_3 = x < -1 cup x > 1.5, G_4 = -1 < x < 1.5$


3. $G_1=A, G_2 = B cup C cup D, G_3 = x < -1, G_4 = x > -1$

---

Where have I gone wrong please?

general-topology complex-analysis

share|cite|improve this question

edited Aug 5 at 11:51

asked Jul 29 at 11:52

BCLC

6,95921973

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.29,30

---

Definitions:

-

---

If I understood the either, or, and, etc right, then $G=(-2,-1) cup D[0,1] cup 1,2$ and thus

• 1.29 (a) $G$ is from left to right, an interval in $mathbb R$, the unit disc, a point in $mathbb R$ and then another point in $mathbb R$.


• 1.29 (b) Interior: $D[0,1]$


• 1.29 (c) Boundary: $[-2,-1] cup C[0,1] cup 1,2 = [-2,-1) cup C[0,1] cup 2$


• 1.29 (d) Isolated: $2$

1.30

Let $A=(-2,-1), B=D[0,1],C=1,D=2$. I'm gonna group these into $G_1$ and $G_2$ separated s.t. $G_1 subseteq G_3$ and $G_2 subseteq G_4$

Ways:

1. $G_1=A cup B cup C, G_2 = D, G_3 = x < 1.5, G_4 = x > 1.5$


2. $G_1=A cup D, G_2 = B cup C, G_3 = x < -1 cup x > 1.5, G_4 = -1 < x < 1.5$


3. $G_1=A, G_2 = B cup C cup D, G_3 = x < -1, G_4 = x > -1$

---

Where have I gone wrong please?

general-topology complex-analysis

share|cite|improve this question

edited Aug 5 at 11:51

asked Jul 29 at 11:52

BCLC

6,95921973

share|cite|improve this question

share|cite|improve this question

edited Aug 5 at 11:51

edited Aug 5 at 11:51

edited Aug 5 at 11:51

asked Jul 29 at 11:52

BCLC

6,95921973

asked Jul 29 at 11:52

BCLC

6,95921973

asked Jul 29 at 11:52

BCLC

6,95921973

6,95921973

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did they really write $-2<z<-1$? I'd get another book....
  – Lord Shark the Unknown
  Jul 29 at 11:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @LordSharktheUnknown Why? It's a perfectly valid property of $z in mathbbC$ to say that $z$ is real and $-2 < z < -1$.
  – Adayah
  Jul 29 at 11:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Two places. First, note a boundary point of G need not be a member of G. Look again at -2. Second, none of your separations split B and C. That's correct, but suggests you should look again at the isolation of 1.
  – David Hartley
  Jul 29 at 12:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidHartley Thanks! ^-^ I actually noticed those transcribed incorrectly. Post as answer?
  – BCLC
  Jul 29 at 12:17
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What does the notation $D[0,1]$ mean?
  – Henning Makholm
  Jul 29 at 13:40
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did they really write $-2<z<-1$? I'd get another book....
  – Lord Shark the Unknown
  Jul 29 at 11:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @LordSharktheUnknown Why? It's a perfectly valid property of $z in mathbbC$ to say that $z$ is real and $-2 < z < -1$.
  – Adayah
  Jul 29 at 11:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Two places. First, note a boundary point of G need not be a member of G. Look again at -2. Second, none of your separations split B and C. That's correct, but suggests you should look again at the isolation of 1.
  – David Hartley
  Jul 29 at 12:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidHartley Thanks! ^-^ I actually noticed those transcribed incorrectly. Post as answer?
  – BCLC
  Jul 29 at 12:17
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What does the notation $D[0,1]$ mean?
  – Henning Makholm
  Jul 29 at 13:40
  

  
  
  
  

Did they really write $-2<z<-1$? I'd get another book....
– Lord Shark the Unknown
Jul 29 at 11:56

Did they really write $-2<z<-1$? I'd get another book....
– Lord Shark the Unknown
Jul 29 at 11:56

1

1

@LordSharktheUnknown Why? It's a perfectly valid property of $z in mathbbC$ to say that $z$ is real and $-2 < z < -1$.
– Adayah
Jul 29 at 11:58

@LordSharktheUnknown Why? It's a perfectly valid property of $z in mathbbC$ to say that $z$ is real and $-2 < z < -1$.
– Adayah
Jul 29 at 11:58

1

1

Two places. First, note a boundary point of G need not be a member of G. Look again at -2. Second, none of your separations split B and C. That's correct, but suggests you should look again at the isolation of 1.
– David Hartley
Jul 29 at 12:15

Two places. First, note a boundary point of G need not be a member of G. Look again at -2. Second, none of your separations split B and C. That's correct, but suggests you should look again at the isolation of 1.
– David Hartley
Jul 29 at 12:15

@DavidHartley Thanks! ^-^ I actually noticed those transcribed incorrectly. Post as answer?
– BCLC
Jul 29 at 12:17

@DavidHartley Thanks! ^-^ I actually noticed those transcribed incorrectly. Post as answer?
– BCLC
Jul 29 at 12:17

1

1

What does the notation $D[0,1]$ mean?
– Henning Makholm
Jul 29 at 13:40

What does the notation $D[0,1]$ mean?
– Henning Makholm
Jul 29 at 13:40

 | 
show 2 more comments

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

Two places. First, note a boundary point of G need not be a member of G. Look again at -2. Second, none of your separations split B and C. That's correct, but suggests you should look again at the isolation of 1. – David Hartley

@DavidHartley Thanks! ^-^ I actually noticed those transcribed incorrectly. Post as answer? – BCLC

share|cite|improve this answer

answered Aug 1 at 5:13

BCLC

6,95921973

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866013%2ftopology-of-2-1-cup-d0-1-cup-1-2%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Two places. First, note a boundary point of G need not be a member of G. Look again at -2. Second, none of your separations split B and C. That's correct, but suggests you should look again at the isolation of 1. – David Hartley

@DavidHartley Thanks! ^-^ I actually noticed those transcribed incorrectly. Post as answer? – BCLC

share|cite|improve this answer

answered Aug 1 at 5:13

BCLC

6,95921973

add a comment | 

up vote
0
down vote



accepted

Two places. First, note a boundary point of G need not be a member of G. Look again at -2. Second, none of your separations split B and C. That's correct, but suggests you should look again at the isolation of 1. – David Hartley

@DavidHartley Thanks! ^-^ I actually noticed those transcribed incorrectly. Post as answer? – BCLC

share|cite|improve this answer

answered Aug 1 at 5:13

BCLC

6,95921973

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Two places. First, note a boundary point of G need not be a member of G. Look again at -2. Second, none of your separations split B and C. That's correct, but suggests you should look again at the isolation of 1. – David Hartley

@DavidHartley Thanks! ^-^ I actually noticed those transcribed incorrectly. Post as answer? – BCLC

share|cite|improve this answer

answered Aug 1 at 5:13

BCLC

6,95921973

Two places. First, note a boundary point of G need not be a member of G. Look again at -2. Second, none of your separations split B and C. That's correct, but suggests you should look again at the isolation of 1. – David Hartley

@DavidHartley Thanks! ^-^ I actually noticed those transcribed incorrectly. Post as answer? – BCLC

share|cite|improve this answer

answered Aug 1 at 5:13

BCLC

6,95921973

share|cite|improve this answer

share|cite|improve this answer

answered Aug 1 at 5:13

BCLC

6,95921973

answered Aug 1 at 5:13

BCLC

6,95921973

answered Aug 1 at 5:13

BCLC

6,95921973

6,95921973

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866013%2ftopology-of-2-1-cup-d0-1-cup-1-2%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email