Difference between revisions of "Open ball"
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| + | ==Definition== | ||
For a [[Metric space|metric space]] <math>(X,d)</math> an "open ball" of radius <math>r</math> centred at <math>a</math> is the set | For a [[Metric space|metric space]] <math>(X,d)</math> an "open ball" of radius <math>r</math> centred at <math>a</math> is the set | ||
<math>\{x\in X|d(a,x)\lt r\}</math>, it can be denoted several ways. I frequently encounter | <math>\{x\in X|d(a,x)\lt r\}</math>, it can be denoted several ways. I frequently encounter | ||
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==Proof that an open ball is open== | ==Proof that an open ball is open== | ||
| − | + | Take the open ball <math>B_\epsilon(p)</math>. | |
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| + | Let <math>x\in B_\epsilon(p)</math> be arbitrary | ||
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| + | Choose <math>r=\epsilon-d(x,p)</math> - then as <math>x\in B_\epsilon(p)\iff d(x,p)<\epsilon</math> we see <math>r>0</math> | ||
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| + | We now need to show that <math>B_r(x)\subset B_\epsilon(p)</math> using the [[Implies and subset relation]] we see: | ||
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| + | <math>B_r(x)\subset B_\epsilon(p)</math><math>\iff y\in B_r(x)\implies y\in B_\epsilon(p)</math> | ||
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| + | So let <math>y\in B_r(x)</math> be arbitrary, then: | ||
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| + | <math>y\in B_r(x)\iff d(y,x)< r=\epsilon-d(x,p)</math> so <math>d(y,x)<\epsilon-d(x,p)</math> | ||
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| + | <math>d(y,x)<\epsilon-d(x,p)\iff d(y,x)+d(x,p)<\epsilon</math> | ||
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| + | But by the [[Triangle inequality]] part of [[Metric space|the metric]] <math>d(y,p)\le d(y,x)+d(x,p)<\epsilon</math> | ||
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| + | So <math>d(y,p)<\epsilon\iff y\in B_\epsilon(p)</math> | ||
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| + | We have shown that <math>y\in B_r(x)\implies y\in B_\epsilon(p)\iff B_r(x)\subset B_\epsilon(p)</math>, since <math>x\in B_\epsilon(p)</math> was arbitrary, we have shown that <math>B_\epsilon(p)</math> is a neighbourhood to all of its points, thus is open. | ||
==See Also== | ==See Also== | ||
* [[Open set]] | * [[Open set]] | ||
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| + | {{Definition|Topology|Metric Space}} | ||
Revision as of 00:19, 9 March 2015
Definition
For a metric space [math](X,d)[/math] an "open ball" of radius [math]r[/math] centred at [math]a[/math] is the set [math]\{x\in X|d(a,x)\lt r\}[/math], it can be denoted several ways. I frequently encounter
[math]B_r(a)=B(a;r)=\{x\in X|d(a,x)\lt r\}[/math] and use [math]B_r(a)[/math]
Proof that an open ball is open
Take the open ball [math]B_\epsilon(p)[/math].
Let [math]x\in B_\epsilon(p)[/math] be arbitrary
Choose [math]r=\epsilon-d(x,p)[/math] - then as [math]x\in B_\epsilon(p)\iff d(x,p)<\epsilon[/math] we see [math]r>0[/math]
We now need to show that [math]B_r(x)\subset B_\epsilon(p)[/math] using the Implies and subset relation we see:
[math]B_r(x)\subset B_\epsilon(p)[/math][math]\iff y\in B_r(x)\implies y\in B_\epsilon(p)[/math]
So let [math]y\in B_r(x)[/math] be arbitrary, then:
[math]y\in B_r(x)\iff d(y,x)< r=\epsilon-d(x,p)[/math] so [math]d(y,x)<\epsilon-d(x,p)[/math]
[math]d(y,x)<\epsilon-d(x,p)\iff d(y,x)+d(x,p)<\epsilon[/math]
But by the Triangle inequality part of the metric [math]d(y,p)\le d(y,x)+d(x,p)<\epsilon[/math]
So [math]d(y,p)<\epsilon\iff y\in B_\epsilon(p)[/math]
We have shown that [math]y\in B_r(x)\implies y\in B_\epsilon(p)\iff B_r(x)\subset B_\epsilon(p)[/math], since [math]x\in B_\epsilon(p)[/math] was arbitrary, we have shown that [math]B_\epsilon(p)[/math] is a neighbourhood to all of its points, thus is open.
