Difference between revisions of "File:R2 top group preimage of 1 to -1 epsilon 0.25 steps 0.125.gif"
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| + | This is an example image for showing that {{M|(\mathbb{R}-\{0\},\times)}} is a [[topological group]]. Consider {{M|m:(\mathbb{R}-\{0\})\times(\mathbb{R}-\{0\})\rightarrow\mathbb{R}-\{0\} }} given by {{M|m:(x,y)\mapsto xy}}, as the example shows {{M|m}} is [[continuous]] (when {{M|(\mathbb{R}-\{0\})^2 }} is considered with the [[product topology]]), an important step in doing that is showing that: | ||
| + | * Given an [[open ball]], {{M|1=B_\epsilon(c)}} which is, in this case, the [[open interval]] {{M|(c-\epsilon,c+\epsilon)}} we must show that the preimage is open (among other things of course). | ||
| + | This image shows the preimage of {{M|(c-\epsilon,c+\epsilon)}} as {{M|c}} goes from {{M|-1}} to {{M|1}} in steps of {{M|0.125}}. For {{M|1=\epsilon:=0.25}}. | ||
| + | You can see for {{M|1=c=0}} the image is symmetric about the {{M|x}}-axis, {{M|y}}-axis, {{M|1=y=x}} line and {{M|1=y=-x}} line and forms like a 4-point star shape. This ''technically'' isn't in the space we're dealing with, however we do see the preimage of {{M|(-0.25,0.25)}} in the space. | ||
Latest revision as of 03:20, 8 August 2016
This is an example image for showing that [ilmath](\mathbb{R}-\{0\},\times)[/ilmath] is a topological group. Consider [ilmath]m:(\mathbb{R}-\{0\})\times(\mathbb{R}-\{0\})\rightarrow\mathbb{R}-\{0\} [/ilmath] given by [ilmath]m:(x,y)\mapsto xy[/ilmath], as the example shows [ilmath]m[/ilmath] is continuous (when [ilmath](\mathbb{R}-\{0\})^2 [/ilmath] is considered with the product topology), an important step in doing that is showing that:
- Given an open ball, [ilmath]B_\epsilon(c)[/ilmath] which is, in this case, the open interval [ilmath](c-\epsilon,c+\epsilon)[/ilmath] we must show that the preimage is open (among other things of course).
This image shows the preimage of [ilmath](c-\epsilon,c+\epsilon)[/ilmath] as [ilmath]c[/ilmath] goes from [ilmath]-1[/ilmath] to [ilmath]1[/ilmath] in steps of [ilmath]0.125[/ilmath]. For [ilmath]\epsilon:=0.25[/ilmath].
You can see for [ilmath]c=0[/ilmath] the image is symmetric about the [ilmath]x[/ilmath]-axis, [ilmath]y[/ilmath]-axis, [ilmath]y=x[/ilmath] line and [ilmath]y=-x[/ilmath] line and forms like a 4-point star shape. This technically isn't in the space we're dealing with, however we do see the preimage of [ilmath](-0.25,0.25)[/ilmath] in the space.
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| current | 03:11, 8 August 2016 |  | 256 × 128 (48 KB) | Alec (Talk | contribs) |
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