Mixing
Determine whether the set ${(𝑥,𝑦,𝑧):𝑥+𝑦+𝑧=2 }$ is open or closed
$begingroup$
I'm trying to figure out whether this set is open or closed ${(x,y,z):x+y+z=2}$, but I'm having a hard time to figure it out in three dimensions.
Can I say that its complement is open and it is therefore closed?
calculus general-topology
edited Jan 31 at 21:06
egreg
185k1486208
asked Jan 31 at 20:00
KarenKaren
927
$endgroup$
add a comment |
$begingroup$
I'm trying to figure out whether this set is open or closed ${(x,y,z):x+y+z=2}$, but I'm having a hard time to figure it out in three dimensions.
Can I say that its complement is open and it is therefore closed?
calculus general-topology
edited Jan 31 at 21:06
egreg
185k1486208
asked Jan 31 at 20:00
KarenKaren
927
$endgroup$
$begingroup$
Is is $x+y+z>2$ or $x+y+z=2$?
$endgroup$
– Bernard
Jan 31 at 20:07
$begingroup$
@Bernad It's $ x+y+z=2$
$endgroup$
– Karen
Jan 31 at 20:08
add a comment |
$begingroup$
I'm trying to figure out whether this set is open or closed ${(x,y,z):x+y+z=2}$, but I'm having a hard time to figure it out in three dimensions.
Can I say that its complement is open and it is therefore closed?
calculus general-topology
edited Jan 31 at 21:06
egreg
185k1486208
asked Jan 31 at 20:00
KarenKaren
927
$endgroup$
I'm trying to figure out whether this set is open or closed ${(x,y,z):x+y+z=2}$, but I'm having a hard time to figure it out in three dimensions.
Can I say that its complement is open and it is therefore closed?
calculus general-topology
calculus general-topology
edited Jan 31 at 21:06
egreg
185k1486208
asked Jan 31 at 20:00
KarenKaren
927
edited Jan 31 at 21:06
egreg
185k1486208
asked Jan 31 at 20:00
KarenKaren
927
edited Jan 31 at 21:06
egreg
185k1486208
edited Jan 31 at 21:06
egreg
185k1486208
edited Jan 31 at 21:06
egreg
185k1486208
185k1486208
asked Jan 31 at 20:00
KarenKaren
927
asked Jan 31 at 20:00
KarenKaren
927
asked Jan 31 at 20:00
KarenKaren
927
927
$begingroup$
Is is $x+y+z>2$ or $x+y+z=2$?
$endgroup$
– Bernard
Jan 31 at 20:07
$begingroup$
@Bernad It's $ x+y+z=2$
$endgroup$
– Karen
Jan 31 at 20:08
add a comment |
$begingroup$
Is is $x+y+z>2$ or $x+y+z=2$?
$endgroup$
– Bernard
Jan 31 at 20:07
$begingroup$
@Bernad It's $ x+y+z=2$
$endgroup$
– Karen
Jan 31 at 20:08
$begingroup$
Is is $x+y+z>2$ or $x+y+z=2$?
$endgroup$
– Bernard
Jan 31 at 20:07
$begingroup$
Is is $x+y+z>2$ or $x+y+z=2$?
$endgroup$
– Bernard
Jan 31 at 20:07
$begingroup$
@Bernad It's $ x+y+z=2$
$endgroup$
– Karen
Jan 31 at 20:08
$begingroup$
@Bernad It's $ x+y+z=2$
$endgroup$
– Karen
Jan 31 at 20:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
Show that any point $(x,y,z)$ in the set $;A=bigl{(x,y,z)mid x+y+zne 2bigr};$ has a small neighbourhood contained in the set. Two cases: either $x+y+z>2$, or $x+y+z<2$.
What can you conclude from this property?
answered Jan 31 at 20:03
BernardBernard
124k741117
$endgroup$
$begingroup$
Off-Topic. $mathrmTeX-mathrmLaTeX$ and Math! Yay!!
$endgroup$
– manooooh
Jan 31 at 20:05
1
$begingroup$
@manooooh: ??? What's off-topic?
$endgroup$
– Bernard
Jan 31 at 20:09
$begingroup$
The comment. It is unrelated to the OP's question.
$endgroup$
– manooooh
Jan 31 at 20:10
$begingroup$
@Bernard by showing so can we conclude that since A is the complement of the set above and A is open, then the set above is closed?
$endgroup$
– Karen
Jan 31 at 20:14
$begingroup$
Yes, exactly${}$.
$endgroup$
– Bernard
Jan 31 at 20:17
add a comment |
$begingroup$
Also, letting $f:mathbb R^3to mathbb R$ be the continuous function $f(x,y,z)=x+y+z$, then $A=f^{-1}({2})$. The set ${2}$ is closed, and inverse images of closed sets are closed.
answered Jan 31 at 20:18
Mike EarnestMike Earnest
27.3k22152
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Show that any point $(x,y,z)$ in the set $;A=bigl{(x,y,z)mid x+y+zne 2bigr};$ has a small neighbourhood contained in the set. Two cases: either $x+y+z>2$, or $x+y+z<2$.
What can you conclude from this property?
answered Jan 31 at 20:03
BernardBernard
124k741117
$endgroup$
$begingroup$
Off-Topic. $mathrmTeX-mathrmLaTeX$ and Math! Yay!!
$endgroup$
– manooooh
Jan 31 at 20:05
1
$begingroup$
@manooooh: ??? What's off-topic?
$endgroup$
– Bernard
Jan 31 at 20:09
$begingroup$
The comment. It is unrelated to the OP's question.
$endgroup$
– manooooh
Jan 31 at 20:10
$begingroup$
@Bernard by showing so can we conclude that since A is the complement of the set above and A is open, then the set above is closed?
$endgroup$
– Karen
Jan 31 at 20:14
$begingroup$
Yes, exactly${}$.
$endgroup$
– Bernard
Jan 31 at 20:17
add a comment |
$begingroup$
Hint:
Show that any point $(x,y,z)$ in the set $;A=bigl{(x,y,z)mid x+y+zne 2bigr};$ has a small neighbourhood contained in the set. Two cases: either $x+y+z>2$, or $x+y+z<2$.
What can you conclude from this property?
answered Jan 31 at 20:03
BernardBernard
124k741117
$endgroup$
$begingroup$
Off-Topic. $mathrmTeX-mathrmLaTeX$ and Math! Yay!!
$endgroup$
– manooooh
Jan 31 at 20:05
1
$begingroup$
@manooooh: ??? What's off-topic?
$endgroup$
– Bernard
Jan 31 at 20:09
$begingroup$
The comment. It is unrelated to the OP's question.
$endgroup$
– manooooh
Jan 31 at 20:10
$begingroup$
@Bernard by showing so can we conclude that since A is the complement of the set above and A is open, then the set above is closed?
$endgroup$
– Karen
Jan 31 at 20:14
$begingroup$
Yes, exactly${}$.
$endgroup$
– Bernard
Jan 31 at 20:17
add a comment |
$begingroup$
Hint:
Show that any point $(x,y,z)$ in the set $;A=bigl{(x,y,z)mid x+y+zne 2bigr};$ has a small neighbourhood contained in the set. Two cases: either $x+y+z>2$, or $x+y+z<2$.
What can you conclude from this property?
answered Jan 31 at 20:03
BernardBernard
124k741117
$endgroup$
Hint:
Show that any point $(x,y,z)$ in the set $;A=bigl{(x,y,z)mid x+y+zne 2bigr};$ has a small neighbourhood contained in the set. Two cases: either $x+y+z>2$, or $x+y+z<2$.
What can you conclude from this property?
answered Jan 31 at 20:03
BernardBernard
124k741117
edited Jan 31 at 20:12
answered Jan 31 at 20:03
BernardBernard
124k741117
answered Jan 31 at 20:03
BernardBernard
124k741117
answered Jan 31 at 20:03
BernardBernard
124k741117
124k741117
$begingroup$
Off-Topic. $mathrmTeX-mathrmLaTeX$ and Math! Yay!!
$endgroup$
– manooooh
Jan 31 at 20:05
1
$begingroup$
@manooooh: ??? What's off-topic?
$endgroup$
– Bernard
Jan 31 at 20:09
$begingroup$
The comment. It is unrelated to the OP's question.
$endgroup$
– manooooh
Jan 31 at 20:10
$begingroup$
@Bernard by showing so can we conclude that since A is the complement of the set above and A is open, then the set above is closed?
$endgroup$
– Karen
Jan 31 at 20:14
$begingroup$
Yes, exactly${}$.
$endgroup$
– Bernard
Jan 31 at 20:17
add a comment |
$begingroup$
Off-Topic. $mathrmTeX-mathrmLaTeX$ and Math! Yay!!
$endgroup$
– manooooh
Jan 31 at 20:05
1
$begingroup$
@manooooh: ??? What's off-topic?
$endgroup$
– Bernard
Jan 31 at 20:09
$begingroup$
The comment. It is unrelated to the OP's question.
$endgroup$
– manooooh
Jan 31 at 20:10
$begingroup$
@Bernard by showing so can we conclude that since A is the complement of the set above and A is open, then the set above is closed?
$endgroup$
– Karen
Jan 31 at 20:14
$begingroup$
Yes, exactly${}$.
$endgroup$
– Bernard
Jan 31 at 20:17
$begingroup$
Off-Topic. $mathrmTeX-mathrmLaTeX$ and Math! Yay!!
$endgroup$
– manooooh
Jan 31 at 20:05
$begingroup$
Off-Topic. $mathrmTeX-mathrmLaTeX$ and Math! Yay!!
$endgroup$
– manooooh
Jan 31 at 20:05
1
1
$begingroup$
@manooooh: ??? What's off-topic?
$endgroup$
– Bernard
Jan 31 at 20:09
$begingroup$
@manooooh: ??? What's off-topic?
$endgroup$
– Bernard
Jan 31 at 20:09
$begingroup$
The comment. It is unrelated to the OP's question.
$endgroup$
– manooooh
Jan 31 at 20:10
$begingroup$
The comment. It is unrelated to the OP's question.
$endgroup$
– manooooh
Jan 31 at 20:10
$begingroup$
@Bernard by showing so can we conclude that since A is the complement of the set above and A is open, then the set above is closed?
$endgroup$
– Karen
Jan 31 at 20:14
$begingroup$
@Bernard by showing so can we conclude that since A is the complement of the set above and A is open, then the set above is closed?
$endgroup$
– Karen
Jan 31 at 20:14
$begingroup$
Yes, exactly${}$.
$endgroup$
– Bernard
Jan 31 at 20:17
$begingroup$
Yes, exactly${}$.
$endgroup$
– Bernard
Jan 31 at 20:17
add a comment |
$begingroup$
Also, letting $f:mathbb R^3to mathbb R$ be the continuous function $f(x,y,z)=x+y+z$, then $A=f^{-1}({2})$. The set ${2}$ is closed, and inverse images of closed sets are closed.
answered Jan 31 at 20:18
Mike EarnestMike Earnest
27.3k22152
$endgroup$
add a comment |
$begingroup$
Also, letting $f:mathbb R^3to mathbb R$ be the continuous function $f(x,y,z)=x+y+z$, then $A=f^{-1}({2})$. The set ${2}$ is closed, and inverse images of closed sets are closed.
answered Jan 31 at 20:18
Mike EarnestMike Earnest
27.3k22152
$endgroup$
add a comment |
$begingroup$
Also, letting $f:mathbb R^3to mathbb R$ be the continuous function $f(x,y,z)=x+y+z$, then $A=f^{-1}({2})$. The set ${2}$ is closed, and inverse images of closed sets are closed.
answered Jan 31 at 20:18
Mike EarnestMike Earnest
27.3k22152
$endgroup$
Also, letting $f:mathbb R^3to mathbb R$ be the continuous function $f(x,y,z)=x+y+z$, then $A=f^{-1}({2})$. The set ${2}$ is closed, and inverse images of closed sets are closed.
answered Jan 31 at 20:18
Mike EarnestMike Earnest
27.3k22152
answered Jan 31 at 20:18
Mike EarnestMike Earnest
27.3k22152
answered Jan 31 at 20:18
Mike EarnestMike Earnest
27.3k22152
answered Jan 31 at 20:18
Mike EarnestMike Earnest
27.3k22152
27.3k22152
add a comment |
add a comment |
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$begingroup$
Is is $x+y+z>2$ or $x+y+z=2$?
$endgroup$
– Bernard
Jan 31 at 20:07
$begingroup$
@Bernad It's $ x+y+z=2$
$endgroup$
– Karen
Jan 31 at 20:08