shah_nosrat Posted March 10, 2012 Share Posted March 10, 2012 Hi, This is the question that needs a proofs, as follows: Show that the smallest element of a nonempty subset of [math]\mathbf{W}[/math] is unique. My attempt at the proof, as follows: Let [math]\mathbf{U} \subseteq \mathbf{W}[/math], by the well ordering principle (WOP) we have that [math]a \in \mathbf{U}[/math] such that [math]a \leq x [/math] [math]\forall x \in \mathbf{U}[/math]. Now suppose [math]b \in \mathbf{U}[/math] such that [math]b \leq x [/math] [math]\forall x \in \mathbf{U}[/math]. Since [math]0 \leq x - a[/math] and [math]0 \leq x - b[/math] by definition. Now, [math]0 \leq x + x - (a + b)[/math] [math]a+ b \leq x + x = 1\cdot x + 1\cdot x = (1+ 1)\cdot x = 2x[/math] [math]a+ b \leq 2x[/math] [math]\frac{a + b}{2} \leq x[/math] . The only way that this inequality will hold [math] \forall x \in \mathbf{U}[/math] is when [math]a = b[/math] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Is the above reasoning and proof correct? Link to comment Share on other sites More sharing options...

Xittenn Posted March 10, 2012 Share Posted March 10, 2012 (edited) If [math] x [/math] is the set of all members of U such that [math] a [/math] is the smallest member of the set then your initial argument or conditions make no sense. If [math] a [/math] is a member of U such that [math] a [/math] is not a member of [math] x [/math] and is smaller than all members of the set [math] x [/math] then I don't see why you couldn't have a [math] b [/math] that meets this same logical condition while not being equal to [math] a [/math]. The proofs I've seen invoke relations where the mapping of a smallest number can hold only for one member or they are equal. It's very picky . . . . I'm sure someone with a little more understanding will enlighten you or me, in either case, after I post this! ** as in there is a relation R that well orders the set, what happens . . . . Edited March 10, 2012 by Xittenn Link to comment Share on other sites More sharing options...

DrRocket Posted March 10, 2012 Share Posted March 10, 2012 Hi, This is the question that needs a proofs, as follows: Show that the smallest element of a nonempty subset of [math]\mathbf{W}[/math] is unique. My attempt at the proof, as follows: Let [math]\mathbf{U} \subseteq \mathbf{W}[/math], by the well ordering principle (WOP) we have that [math]a \in \mathbf{U}[/math] such that [math]a \leq x [/math] [math]\forall x \in \mathbf{U}[/math]. Now suppose [math]b \in \mathbf{U}[/math] such that [math]b \leq x [/math] [math]\forall x \in \mathbf{U}[/math]. Since [math]0 \leq x - a[/math] and [math]0 \leq x - b[/math] by definition. Now, [math]0 \leq x + x - (a + b)[/math] [math]a+ b \leq x + x = 1\cdot x + 1\cdot x = (1+ 1)\cdot x = 2x[/math] [math]a+ b \leq 2x[/math] [math]\frac{a + b}{2} \leq x[/math] . The only way that this inequality will hold [math] \forall x \in \mathbf{U}[/math] is when [math]a = b[/math] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Is the above reasoning and proof correct? This theorem has nothing to do with well-ordering. It only requires linear ordering. The point is that no set can have two or more "least elements". You only need well-ordering in order to conclude that a non-empty set has at least one least element, but existence of a least element is not required for the theorem as stated. Link to comment Share on other sites More sharing options...

shah_nosrat Posted March 12, 2012 Author Share Posted March 12, 2012 This theorem has nothing to do with well-ordering. It only requires linear ordering. The point is that no set can have two or more "least elements". You only need well-ordering in order to conclude that a non-empty set has at least one least element, but existence of a least element is not required for the theorem as stated. Okay, taking your advice. Forget about my previous attempt at the proof. Using Linear (total) ordering. If we take [math]\mathbf{U} \subseteq W[/math] ---> I need to invoke existence for it to make sense (To me anyway) suppose [math]a, b \in \mathbf{U}[/math] with the property of being the least elements for all elements in [math]\mathbf{U}[/math]. Now, because of the antisymmetry property of the linear ordering then, as follows: if aRb and bRa then a = b ---> Does this conclude the least element in the subset if it exist is unique? Your help is once again appreciated Link to comment Share on other sites More sharing options...

DrRocket Posted March 12, 2012 Share Posted March 12, 2012 Now, because of the antisymmetry property of the linear ordering then, as follows: if aRb and bRa then a = b ---> Does this conclude the least element in the subset if it exist is unique? Yes, it really is just this simple. If as least element exist it is unique simply because any two least elements are equal. 1 Link to comment Share on other sites More sharing options...

shah_nosrat Posted March 12, 2012 Author Share Posted March 12, 2012 Yes, it really is just this simple. If as least element exist it is unique simply because any two least elements are equal. Thank you Dr.Rocket. You are giving me confidence in art of proofing. Yes, it really is just this simple. If as least element exist it is unique simply because any two least elements are equal. Thank you Dr.Rocket. You are giving me confidence in art of proofing. Link to comment Share on other sites More sharing options...

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