Difference between revisions of "Open ball"

Revision as of 20:51, 12 February 2015 (view source) Alec (Talk | contribs) m ← Older edit		Revision as of 00:19, 9 March 2015 (view source) Alec (Talk | contribs) m Newer edit →
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−	{{Definition|Topology|Metric Space}}
		+	==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==
−	{{Todo|The proof is really easy, just show a smaller ball fits inside, thus the open ball is a neighborhood to all of its points}}	+	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 [[Metric space|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.
	==See Also==		==See Also==
	* [[Open set]]		* [[Open set]]
		+
		+	{{Definition|Topology|Metric Space}}

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Revision as of 00:19, 9 March 2015

Definition

For a metric space

$$(X,d)$$ an "open ball" of radius $$r$$ centred at $$a$$ is the set $$\{x\in X|d(a,x)\lt r\}$$ , it can be denoted several ways. I frequently encounter

$$B_r(a)=B(a;r)=\{x\in X|d(a,x)\lt r\}$$ and use $$B_r(a)$$

Proof that an open ball is open

Take the open ball

$$B_\epsilon(p)$$ .

Let

$$x\in B_\epsilon(p)$$ be arbitrary

Choose

$$r=\epsilon-d(x,p)$$ - then as $$x\in B_\epsilon(p)\iff d(x,p)<\epsilon$$ we see $$r>0$$

We now need to show that

$$B_r(x)\subset B_\epsilon(p)$$ using the Implies and subset relation we see:

$$B_r(x)\subset B_\epsilon(p)$$ $$\iff y\in B_r(x)\implies y\in B_\epsilon(p)$$

So let

$$y\in B_r(x)$$ be arbitrary, then:

$$y\in B_r(x)\iff d(y,x)< r=\epsilon-d(x,p)$$ so $$d(y,x)<\epsilon-d(x,p)$$

$$d(y,x)<\epsilon-d(x,p)\iff d(y,x)+d(x,p)<\epsilon$$

But by the Triangle inequality part of the metric

$$d(y,p)\le d(y,x)+d(x,p)<\epsilon$$

So

$$d(y,p)<\epsilon\iff y\in B_\epsilon(p)$$

We have shown that

$$y\in B_r(x)\implies y\in B_\epsilon(p)\iff B_r(x)\subset B_\epsilon(p)$$ , since $$x\in B_\epsilon(p)$$ was arbitrary, we have shown that $$B_\epsilon(p)$$ is a neighbourhood to all of its points, thus is open.

See Also

• Open set