Metric Spaces | Lecture 25 | Subset of Discrete Metric Space is always Open

[@kktech04]

What a nice explanation!

[@kevizekhorhetso6189]

Great explanation!!👍

[@chiranjeevkaur7716]

Thank you for clearing this👍

[@manutripathi8450]

Very happy to see your lectures sir it's really helped me a lot thank you so much sir

[@ranjankhatu]

Most welcome

[@pramodverma406]

😊great explain

[@shaapanwesley4167]

Nice explanation I get it right

[@adebayoemmanuel911]

God bless you Sir

[@ranjankhatu]

Thank you

[@rsg4666]

Thank you sir

[@user-pd4fd5hh2c]

❤❤❤

[@tushar9605]

If r>1 not 1/2 then open ball(or sphere ) will be whole metric space X which is not contained in any arbitrary set ..so subset of metric space will not be open ..?Sir please clarify my doubt

[@ranjankhatu]

In definition of open set, we say, there exists r > 0, but not for every r > 0

[@ranjankhatu]

Therefore it is enough to find atleast one r which will satisfy the condition

[@tushar9605]

@@ranjankhatu ohh ..now I understood .sir thanks for clearing my doubt .

[@sachinpundir6185]

Nicely explained

[@ranjankhatu]

Thank you

[@kumardeepmukhopadhyay7393]

I think y should be an element of subset G , not the metric space X because x is an element of G and we are proving that the open ball about x is a subset of G.

[@ranjankhatu]

@@kumardeepmukhopadhyay7393 no. Simply I am following the definition of Open ball.

[@322-pinkidhobi9]

Consider (Z, | |) as a subspace of (R, | |), then open balls in Z are (a) infinite sets (b) singleton sets (c) finite sets (d) None of these

[@322-pinkidhobi9]

.

[@ranjankhatu]

Finite sets

[@322-pinkidhobi9]

Sir please explain

[@Mrcrazyindian777]

Sir jaruri nhi hai ki english me bolkar padhaye aap..aap accha padha rhe hai but sir use hindi language also because sir acche se samjh me bhi aana chahiye n tabhi sir padhne me accha lagta hai

[@shaapanwesley4167]

Check email please I contacted you 🙏

[@ranjankhatu]

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