Difference between revisions of "Characteristic property of the subspace topology"
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==Proof== | ==Proof== | ||
− | {{Requires proof|grade=A|msg=Already done on task page, just copy and paste!}} | + | {{Requires proof|grade=A|msg=Already done on task page, just copy and paste! '''SEE [[Task:Characteristic property of the subspace topology|HERE]] PROOF IS ALREADY DONE'''}} |
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==See also== | ==See also== | ||
{{Todo|Put some links here}} | {{Todo|Put some links here}} |
Latest revision as of 23:25, 25 September 2016
Stub grade: A
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
I've already written the proof in a task page so it should be easy to put this all together
Contents
Statement
- Given any topological space [ilmath](Y,\mathcal{ K })[/ilmath] and any map [ilmath]f:Y\rightarrow S[/ilmath] we have:
- [ilmath](f:Y\rightarrow S [/ilmath] is continuous[ilmath])\iff(i_S\circ f:Y\rightarrow X [/ilmath] is continuous[ilmath])[/ilmath]
Where [ilmath]i_S:S\rightarrow X[/ilmath] given by [ilmath]i_S:s\mapsto s[/ilmath] is the canonical injection of the subspace topology (which is itself continuous)[Note 2]
Proof
Grade: A
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
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The message provided is:
The message provided is:
Already done on task page, just copy and paste! SEE HERE PROOF IS ALREADY DONE
See also
TODO: Put some links here
Notes
- ↑ This means [ilmath]S\in\mathcal{P}(X)[/ilmath], or [ilmath]S\subseteq X[/ilmath] of course
- ↑ This leads to two ways to prove the statement:
- If we show [ilmath]i_S:S\rightarrow X[/ilmath] is continuous, then we can use the composition of continuous maps is continuous to show if [ilmath]f[/ilmath] continuous then so is [ilmath]i_S\circ f[/ilmath]
- We can show the property the "long way" and then show [ilmath]i_S:S\rightarrow X[/ilmath] is continuous as a corollary
References
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