Proving one interval is a subset of others

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I am trying to prove that $left(a,bright)$ is an open set in $mathbbR$ under the usual metric $dleft(x,yright)=|x-y|$
given any $xin left(a,bright)$ I am supposed to produce an open ball which is contained in $left(a,bright)$.
I think $epsilon=minx-a,b-x$ will work for us as the radius of the ball. That is $left(x-epsilon,x+epsilonright) subset left(a,bright)$
I am unable to prove the following mathematically.
$left(x-epsilon,x+epsilonright) subset left(a,bright)$
Help me with this.
real-analysis
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up vote
2
down vote
favorite
I am trying to prove that $left(a,bright)$ is an open set in $mathbbR$ under the usual metric $dleft(x,yright)=|x-y|$
given any $xin left(a,bright)$ I am supposed to produce an open ball which is contained in $left(a,bright)$.
I think $epsilon=minx-a,b-x$ will work for us as the radius of the ball. That is $left(x-epsilon,x+epsilonright) subset left(a,bright)$
I am unable to prove the following mathematically.
$left(x-epsilon,x+epsilonright) subset left(a,bright)$
Help me with this.
real-analysis
The desired containment is equivalent to $a leq x - epsilon$ and $x + epsilon leq b$, and these in turn follow immediately from your definition of $epsilon$.
â Bungo
Sep 8 at 5:50
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am trying to prove that $left(a,bright)$ is an open set in $mathbbR$ under the usual metric $dleft(x,yright)=|x-y|$
given any $xin left(a,bright)$ I am supposed to produce an open ball which is contained in $left(a,bright)$.
I think $epsilon=minx-a,b-x$ will work for us as the radius of the ball. That is $left(x-epsilon,x+epsilonright) subset left(a,bright)$
I am unable to prove the following mathematically.
$left(x-epsilon,x+epsilonright) subset left(a,bright)$
Help me with this.
real-analysis
I am trying to prove that $left(a,bright)$ is an open set in $mathbbR$ under the usual metric $dleft(x,yright)=|x-y|$
given any $xin left(a,bright)$ I am supposed to produce an open ball which is contained in $left(a,bright)$.
I think $epsilon=minx-a,b-x$ will work for us as the radius of the ball. That is $left(x-epsilon,x+epsilonright) subset left(a,bright)$
I am unable to prove the following mathematically.
$left(x-epsilon,x+epsilonright) subset left(a,bright)$
Help me with this.
real-analysis
real-analysis
asked Sep 8 at 5:16
user581912
856
856
The desired containment is equivalent to $a leq x - epsilon$ and $x + epsilon leq b$, and these in turn follow immediately from your definition of $epsilon$.
â Bungo
Sep 8 at 5:50
add a comment |Â
The desired containment is equivalent to $a leq x - epsilon$ and $x + epsilon leq b$, and these in turn follow immediately from your definition of $epsilon$.
â Bungo
Sep 8 at 5:50
The desired containment is equivalent to $a leq x - epsilon$ and $x + epsilon leq b$, and these in turn follow immediately from your definition of $epsilon$.
â Bungo
Sep 8 at 5:50
The desired containment is equivalent to $a leq x - epsilon$ and $x + epsilon leq b$, and these in turn follow immediately from your definition of $epsilon$.
â Bungo
Sep 8 at 5:50
add a comment |Â
1 Answer
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Maybe this small picture will help a bit.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote

Maybe this small picture will help a bit.
add a comment |Â
up vote
0
down vote

Maybe this small picture will help a bit.
add a comment |Â
up vote
0
down vote
up vote
0
down vote

Maybe this small picture will help a bit.

Maybe this small picture will help a bit.
answered Sep 8 at 6:08
dmtri
956518
956518
add a comment |Â
add a comment |Â
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The desired containment is equivalent to $a leq x - epsilon$ and $x + epsilon leq b$, and these in turn follow immediately from your definition of $epsilon$.
â Bungo
Sep 8 at 5:50