prove that null set is a subset of every set.?
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Well if you make E (everything possible) value or setwise then the null set is included in everything:\n   \n∅ ⊆ E\n\nThen for any set or subset A in E you would have either\n\n∅ ⊂ A   or   ∅ ⊂ A'     since it has to be in one of these since the complement of A, which is A' is everything else in E not in A.  This is obvious.\n\nLet's assume the second always, that the null set is a subset of the complement of any given set and not a part of set we look at.  Then for A the nullset is located in A'\n\nBut then let's look at a set B which is A', and if we follow our rule we assume then the nullset is located in B' which is A.  Then from this method we can see that the nullset is a subset of both A and B and A' and B'.  It is obvious from this point that any given set has ∅ as a subset.\n\nQ.E.D.
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conditional statement and its inverse
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So X is dependent on TSOUT (the source of universal truth)\n\nAnd the TSOUT is dependent on X.\n\nThat, my friend, is circular logic, and as such, is a false proof.
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not sure ...but may be u have to do like \n\nif (conn!=null)\n   connected=true;\nelse\n   connected=false;\n\nNot sure abt this...for more info google it
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{x: x inequal to 2}\nor\n{x: x=/= 2}
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if B' is subset of A' then\n\nif x belongs to B' then x belongs to A'\n\nif x belongs to A then x cannot belong to B' because if x belongs to B' then it belongs to A then x belongs to A and A'\n\nso if  x belongs to A the x belongs to B(as it does not belong to B')\n\nso A is subset of B
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Any finite collection?\n\nsum of all subcovers is a finite cover.
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1. Set A is subset of set B if and only if any element in A is in B. For example, {2,3,5} is a subset of {1,2,3,4,5}.\n An integer cannot be subset of any set, since itself is not a set.\n\n 2. Composite number is a positive integer, which can be written in product of two positive integers that are bigger than 2. Any positive number other than prime numbers and 1(1 is neither a prime number nor composite number) is composite number.\nFirst few composite numbers are 4,6,8,9,10,12,14,15,16,18,20,...\n\nThe last two questions are not clear. Can you clarify it?
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look at my question\n\nBidDogARRRRR...\n1 minute ago\nDoes anyone know 0=1 and anything = anything else?\na = 1; b = 0\n1. a = b + 1 \n2. (a-b)a = (a-b)(b+1)\n3. a2 - ab = ab + a - b2 - b \n4. a2 - ab -a = ab + a -a - b2 - b\n5. a(a - b - 1) = b(a - b - 1) \n6. a = b\n7. b + 1 = b \n8. Therefore, 1 = 0
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Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
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You need to start from the definintions of x < y and x > y.\n\nThey probably go like this\nif x < y, then there exists 0 < z such that x+z = y.\nif x > y, then there exists 0 < w such that x = y+w.\n\nPartial proof:\nif x < y, and x = y, then for some z, 0 < z, x+z = y.\nsince you have x = y, too, y + z = y, z = 0. That contradicts 0 < z. Proving the other two parts is similar.\nx+z = y,
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I have a copy of the second edition. Is that good enough?\n\nIs there something in particular you need, or do you want the outline of the whole book?\n\n1. Set theory\nIntroduction\nFunctions\nUnions, Intersections, and complements\nAlgebra of sets\nThe axiom of chice and infinite direct product\nCountable sets\nRelations and equivalences\nPartial orderings and the maximal principle\nWell ordering and the countable ordinal\n\nPart One -- Theory of Functions of a Real Variable\n\n2. The Real Number System\n\nAxioms for the real numbers\nThe natural and rational numbers as subsets of R\nThe extended real numbers\nSequences of real numbers\nOpen and closed sets of real numbers\nContinuous functions\nBorel sets\n\n3. Lebesgue Measure\nIntroduction\nOuter measure\nMeasurable sets and Lebesgue measure\nA nonmeasurable set\nMeasurable functions\nLittlwood's three principles\n\n4. The Lebesgue Integral\nThe Riemann integral\nThe Lebesgue integral of a bounded function over a set of finite measure\nThe integral of a nonnegative function\nThe general Lebesque integral\nConvergence in measure\n\n5. Differentiation and Integration\nDifferentiation of a monotone function\nFunction of bounded variation\nDifferentiation of an integral\nAbsolute continuity\nConvex functions\n\n6. The Classical Banach Spaces\nThe Lp spaces\nThe Holder Minkowski inequalities\nConvergence and completeneww\nBounded linear functionsals on the Lp spaces\n\n\nPart Two -- Abstract Spaces\n\n7. Metric Spaces\nIntroduction\nOpen and closed sets\nContinuous function and homeomorphisms\nConvergence and completeness\nUniform continuity and uniformity\nSubspaces\nBaire category\n\n8. Topological Spaces\nFundamental notions\nBases and countability\nThe separation axioms and continuous real-valued functions\nProduct spaces\nConnectedness\nAbsolute (G delta)\nNets\n\n9. Compact Spaces\nBasic properties\nCountable compactness and the Bolzano-Weierstrass property\nCompact metric spaces\nProducts of compact spaces\nLocally compact spaces\nThe Stone-Cech compactification\nThe Stone-Weierstrass theorem\nThe Ascoli theorem\n\n10. Banach Spaces\nIntroduction\nLinear Operators\nLinear Functionals and the Hahn-Banach theorem\nThe closed graph theorem\nTopological vector spaces\nWeak topologies\nConvexity\nHilbert space\n\nPart Three -- General Measure and Integration Theory\n\n11. Measure and Integration\nMeasure spaces\nMeasurable functions\nIntegration\nGeneral convergence theorems\nSigned measures\nThe Radon-Nikodym theorem\nThe Lp spaces\n\n12. Measure and Outer Measure\nOuter measure and measurability\nThe extension theorem\nThe Lebesgue-Stieltjes integral\nProduct measures\nInner measure\nExtension by sets of measure zero\nCaratheodory outer measure\n\n13. The Daniell Integral\nIntroduction\nThe extension theorem\nUniqueness\nMeasureability and measure\n\n14. Measure and Topology\nBaire sets and Borel sets\nPositive linear functionals and Baire measure\nBounded linear fuctionals on C(X)\nThe Borel extension of a measure\n\n15. Mappings of Measure Spaces\nPoint mappings and set mappings\nMeasure algebras\nBorel equivalences\nSet mappings and point mappings ofn complete metric spaces\nThe isometries of Lp
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ok, i think i have an idea for this one. \nSketch of a proof:\nSuppose that a is an element of order k. Also suppose that f is an isomorphism. Then we know that f(a*a*a...k times)=f(a)*f(a)*f(a)*...k times (homomorphism definition). We know that f(a)^k is the identity because isomorphisms take the identity to the identity. Now see that k must be the first number with the quality f(a)^k=e. This comes as a result of the isomorphism having a kernel with one element. If we suppose for the sake of contradiction that f(a)^m=e with m<k, then we would have f(a*a*a*a...m times)=e, so that means that f(a^m)=e at the same time as f(a^k)=e with a^m not equal a^k, this is a contradiction.\n\nQED...ish
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x(abs[x- (x^2)] (-1=2)\n\n:> peace\n.
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There has been no development in this field, but you can check out the following result pertaining to set theory and group theory.\n\nThe power set P(x) of a set X, is a not a group under (U) union operation.\n\nProof : \nTo prove this we try to prove three properties \n1. Associativity \nFor A, B, C are in P(x) then \n(AUB)UC = A U (BUC)  [This is more or less obvious or simple to prove]\n\n2. Identity element\nThere is an an element(e) in P(x) such that for all A in P(x)\ne U A = A U e = A\nAnd, this element for power set is 'empty set' . [Sorry, but I dont have 'phi' symbol here.]\n\n3. Inverse element\nFor each A  in P(x) there exist another element B in P(x) such that\n(AUB) = (BUA) = e(empty set)\n\nThis property is not possible because you can not produce an empty set with union operation with any other set(irrespective wether its empty or not)\n\nSo, its not a group under Union.\n\n\nPlease noe that : The power set P(x) of a set X, is a not a group under intersection operation.
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It is not a group for the following reason.  \n\nThe elements of P(X) are all the subsets of X.\n\nA group must have an identity element, i.e. there is an subset e of X such that if Y is a subset of X then e U Y = Y U e = Y.  Now since the empty set is a subset of every set and since Y U empty set = empty set U Y = Y for any Y then the empty set is the identity element.  \n\nBut each element of a group has an inverse in the group, i.e. for any subset Y of X there is a subset Z of X such that Y U Z = Z U Y = the identity element = the empty set. Consider any non-empty subset of X, for example {a}.  There is no subset Z of X such that {a} U Z = Z U {a} = the empty set since {a} U Z always contains at least a.  \n\n{P(X),U} is not a group since the non-empty subsets of X have no inverses.
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The rationals are a countable subset of [0,1], which\nis an uncountable set. The probability that f(x) = 1 \nis therefore 0.