The result that if two continuous maps agree on a dense subset of the domain, they agree everywhere is really reminiscent of the fact that two linear maps are identical if they agree at a basis. I wonder if there is an underlying analogy here, or perhaps a unifying concept that makes things such as these more transparent.
Only significant result in complete metric spaces is the Baire's theorem. It is published long ago. Other than this, you may need to look at some examples. Refer to my book "Topology of Metric Spaces".
@@kumarhcu sir I am using both of your books functional analysis and Topology of metric space but I get stuck in some examples how to proof them. #Thank you sir
It is true... We can prove that easily... Suppose A is not dense in X then there exists an open ball (say with radius r) in X such that A intersection with that open ball is empty now take the centre of that open ball and fix any y in Y and construct an open ball centered at (x,y) with radius less than r, then this ball B((x,y),r) will have intersection empty with A×B.... (Why?)... So it is a contradiction to out assumption making A×B not dense in X×Y...
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wow what example did you give of dense subset . thank you sir 😊🙏🙏🙏..
Wow, thank you so much.
The result that if two continuous maps agree on a dense subset of the domain, they agree everywhere is really reminiscent of the fact that two linear maps are identical if they agree at a basis. I wonder if there is an underlying analogy here, or perhaps a unifying concept that makes things such as these more transparent.
So appreciable
Sir if possible please complete this upto connectedness.. it's a request
Will be done in near future.
Sir kindly upload a video on complete metric space.
Only significant result in complete metric spaces is the Baire's theorem. It is published long ago. Other than this, you may need to look at some examples. Refer to my book "Topology of Metric Spaces".
@@kumarhcu sir I am using both of your books functional analysis and Topology of metric space but I get stuck in some examples how to proof them. #Thank you sir
Sir functional analysis ki book aapki hai kya
www.amazon.in/FUNCTIONAL-ANALYSIS-First-Course-KUMARESAN/dp/B07VCT6R57
Check it.
Thank you sir.
Yes, if A subset of X and B subset of Y and AxB is dense in XxY ,then A must be dense in X and B must be dense in Y 😊. thank you sir🙏🙏 .am I right ?
It is true... We can prove that easily... Suppose A is not dense in X then there exists an open ball (say with radius r) in X such that A intersection with that open ball is empty now take the centre of that open ball and fix any y in Y and construct an open ball centered at (x,y) with radius less than r, then this ball B((x,y),r) will have intersection empty with A×B.... (Why?)... So it is a contradiction to out assumption making A×B not dense in X×Y...