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Let S be a bounded set and let GLB(S) be its greatest lower bound. A.pdf

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Let S be a bounded set and let GLB(S) be its greatest lower bound. Assume that GLB(S) is not an element of S. Prove that GLB(S) is an acumulation point of S. Solution Let S be a bounded set and let GLB(S) be its greatest lower bound. THAT IS THERE IS NUMBER L , SUCH THAT NO ELEMENT IN THE SET S IS SMALLER THAN L.. FO EX ..IF S = [1,1/2 , 1/3,1/4...] ETC WE CAN SAY L= -1 , SINCE NO ELEMENT IN THE SET IS LESS THAN L=-1 ..WE CAN EVEN SAY L=-0.5 AS A GLB , SINCE THAT IS ALSO TRUE .. Assume that GLB(S) is not an element of S...OK... IT CAN BE FROM THE DEFINITION OF GLB , AS WE HAVE SHOWN ABOVE ... Prove that GLB(S) is an acumulation point of S. SO LET US TAKE ANY SEQUENCE [ S1,S2,...Sn] FROM SET S .... SINCE S IS A BOUNDED SET AS GIVEN IN SOME SPACE SAY RN , THIS SEQUENCE SHALL BE CONTAINED IN A BOUNDED CLOSED BALL & SO IS A COMPACT SUB SET OF THE SPACE RN... THE SEQUENCE THUS HAS A CONVERGING SUBSEQUENCE IN RN AND THE LIMIT OF THAT CONVERGENCE IS THE ACCUMULATION POINT OF A ...

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Let S be a bounded set and let GLB(S) be its greatest lower bound. Assume that GLB(S) is not an element of S. Prove that GLB(S) is an acumulation point of S. Solution Let S be a bounded set and let GLB(S) be its greatest lower bound. THAT IS THERE IS NUMBER L , SUCH THAT NO ELEMENT IN THE SET S IS SMALLER THAN L.. FO EX ..IF S = [1,1/2 , 1/3,1/4...] ETC WE CAN SAY L= -1 , SINCE NO ELEMENT IN THE SET IS LESS THAN L=-1 ..WE CAN EVEN SAY L=-0.5 AS A GLB , SINCE THAT IS ALSO TRUE .. Assume that GLB(S) is not an element of S...OK... IT CAN BE FROM THE DEFINITION OF GLB , AS WE HAVE SHOWN ABOVE ... Prove that GLB(S) is an acumulation point of S. SO LET US TAKE ANY SEQUENCE [ S1,S2,...Sn] FROM SET S .... SINCE S IS A BOUNDED SET AS GIVEN IN SOME SPACE SAY RN , THIS SEQUENCE SHALL BE CONTAINED IN A BOUNDED CLOSED BALL & SO IS A COMPACT SUB SET OF THE SPACE RN... THE SEQUENCE THUS HAS A CONVERGING SUBSEQUENCE IN RN AND THE LIMIT OF THAT CONVERGENCE IS THE ACCUMULATION POINT OF A ..

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Let S be a bounded set and let GLB(S) be its greatest lower bound. A.pdf

  • 1. Let S be a bounded set and let GLB(S) be its greatest lower bound. Assume that GLB(S) is not an element of S. Prove that GLB(S) is an acumulation point of S. Solution Let S be a bounded set and let GLB(S) be its greatest lower bound. THAT IS THERE IS NUMBER L , SUCH THAT NO ELEMENT IN THE SET S IS SMALLER THAN L.. FO EX ..IF S = [1,1/2 , 1/3,1/4...] ETC WE CAN SAY L= -1 , SINCE NO ELEMENT IN THE SET IS LESS THAN L=-1 ..WE CAN EVEN SAY L=-0.5 AS A GLB , SINCE THAT IS ALSO TRUE .. Assume that GLB(S) is not an element of S...OK... IT CAN BE FROM THE DEFINITION OF GLB , AS WE HAVE SHOWN ABOVE ... Prove that GLB(S) is an acumulation point of S. SO LET US TAKE ANY SEQUENCE [ S1,S2,...Sn] FROM SET S .... SINCE S IS A BOUNDED SET AS GIVEN IN SOME SPACE SAY RN , THIS SEQUENCE SHALL BE CONTAINED IN A BOUNDED CLOSED BALL & SO IS A COMPACT SUB SET OF THE SPACE RN... THE SEQUENCE THUS HAS A CONVERGING SUBSEQUENCE IN RN AND THE LIMIT OF THAT CONVERGENCE IS THE ACCUMULATION POINT OF A ..