Metric Spaces | Lecture 47 | Every Singleton Set is a Closed Set

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  • Опубліковано 21 гру 2024

КОМЕНТАРІ • 11

  • @ashishtiwari1060
    @ashishtiwari1060 2 роки тому +2

    Your teaching style is very awsome

  • @letsmakemathssimple5860
    @letsmakemathssimple5860 2 роки тому +3

    Your teaching is fabulous sir

  • @silasarmah602
    @silasarmah602 Рік тому +1

    Nice presentation

  • @ashutosharora2196
    @ashutosharora2196 2 роки тому

    After the step r = d(x,p) > 0 we say that we create an open ball of radius r with center p should be a subset of X - A. I want to ask why have we created a ball of radius r why not K > R? Because if we will create a ball of radius greater than R then the open ball will also contain the singleton set.

    • @ranjankhatu
      @ranjankhatu  2 роки тому +1

      If we take a ball with Radius greater than r then it will contain point p, here our target is to find a ball which does not contain p i.e. it should lie in X-A

  • @asmakotwal6624
    @asmakotwal6624 2 роки тому

    pls make video on compact ms

  • @ashishtiwari1060
    @ashishtiwari1060 2 роки тому

    Sir please upload full bsc 3 rd year mathematics

  • @keshavlalit1555
    @keshavlalit1555 2 роки тому

    A={f€C[a,b] ; f(a)=1 } sir ise closed subset prove krna h C[a,b] ka w.r.t supremum metric.

  • @heartyrhythms1697
    @heartyrhythms1697 Рік тому

    Hi sir,my sister is having supply in this subject,can you pls help her with a tution online.can I have your contact pls.