sets that do not have an infimum

Some examples for sets that do not have an infimum:

•
The set $M_{1}:=\\mathbb{Q}$ (as a subset of $\\mathbb{Q}$ ) does not havean infimum (nor a supremum). Intuitively this is clear, as the set isunbounded. The (easy) formal proof is left as an exercise for the reader.
•
A more interesting example: The set $M_{2}:=\\{x\\in\\mathbb{Q}:x^{2}\\geq 2,x>0\\}$ (again as a subset of $\\mathbb{Q}$ ) .
Proof.
Clearly, $\\inf(M_{2})>0$ . Assume $i>0$ is an infimum of $M_{2}$ . Now we usethe fact that $\\sqrt{2}$ is not rational, and therefore $i<\\sqrt{2}$ or $i>\\sqrt{2}$ .
If $i<\\sqrt{2}$ , choose any $j\\in\\mathbb{Q}$ from the interval $(i,\\sqrt{2})\\subset\\mathbb{R}$ (this is a real interval, but as the rational numbersare dense (http://planetmath.org/Dense) in the real numbers, every nonempty interval in $\\mathbb{R}$ contains a rational number, hence such a $j$ exists).
Then $j>i$ , but $j<\\sqrt{2}$ , hence $j^{2}<2$ and therefore $j$ is a lowerbound for $M_{2}$ , which is a contradiction.
On the other hand, if $i>\\sqrt{2}$ , the argument is very similar:Choose any $j\\in\\mathbb{Q}$ from the interval $(\\sqrt{2},i)\\subset\\mathbb{R}$ . Then $j<i$ , but $j>\\sqrt{2}$ , hence $j^{2}>2$ and therefore $j\\in M_{2}$ . Thus $M_{2}$ contains an element $j$ smaller than $i$ , which isa contradiction to the assumption that $i=\\inf(M_{2})$ ∎
Intuitively speaking, this example exploits the fact that $\\mathbb{Q}$ doesnot have “enough elements”. More formally, $\\mathbb{Q}$ as a metric spaceis not complete (http://planetmath.org/Complete). The $M_{2}$ defined above is the real interval $M_{2}^{\\prime}:=(\\sqrt{2},\\infty)\\subset\\mathbb{R}$ intersected with $\\mathbb{Q}$ . $M_{2}^{\\prime}$ as a subset of $\\mathbb{R}$ does have an infimum (namely $\\sqrt{2}$ ),but as that is not an element of $\\mathbb{Q}$ , $M_{2}$ does not have aninfimum as a subset of $\\mathbb{Q}$ .
This example also makes it clear that it is important to clearly state thesuperset one is working in when using the notion of infimum or supremum.
It also illustrates that the infimum is a natural generalization of theminimum of a set, as a set that does not have a minimum may still havean infimum (such as $M_{2}^{\\prime}$ ).
Of course all the ideas expressed here equally apply to the supremum, as thetwo notions are completely analogous (just reverse all inequalities).

单词	SetsThatDoNotHaveAnInfimum
释义	sets that do not have an infimum Some examples for sets that do not have an infimum: • The set $M_{1}:=\\mathbb{Q}$ (as a subset of $\\mathbb{Q}$ ) does not havean infimum (nor a supremum). Intuitively this is clear, as the set isunbounded. The (easy) formal proof is left as an exercise for the reader. • A more interesting example: The set $M_{2}:=\\{x\\in\\mathbb{Q}:x^{2}\\geq 2,x>0\\}$ (again as a subset of $\\mathbb{Q}$ ) . Proof. Clearly, $\\inf(M_{2})>0$ . Assume $i>0$ is an infimum of $M_{2}$ . Now we usethe fact that $\\sqrt{2}$ is not rational, and therefore $i<\\sqrt{2}$ or $i>\\sqrt{2}$ . If $i<\\sqrt{2}$ , choose any $j\\in\\mathbb{Q}$ from the interval $(i,\\sqrt{2})\\subset\\mathbb{R}$ (this is a real interval, but as the rational numbersare dense (http://planetmath.org/Dense) in the real numbers, every nonempty interval in $\\mathbb{R}$ contains a rational number, hence such a $j$ exists). Then $j>i$ , but $j<\\sqrt{2}$ , hence $j^{2}<2$ and therefore $j$ is a lowerbound for $M_{2}$ , which is a contradiction. On the other hand, if $i>\\sqrt{2}$ , the argument is very similar:Choose any $j\\in\\mathbb{Q}$ from the interval $(\\sqrt{2},i)\\subset\\mathbb{R}$ . Then $j<i$ , but $j>\\sqrt{2}$ , hence $j^{2}>2$ and therefore $j\\in M_{2}$ . Thus $M_{2}$ contains an element $j$ smaller than $i$ , which isa contradiction to the assumption that $i=\\inf(M_{2})$ ∎ Intuitively speaking, this example exploits the fact that $\\mathbb{Q}$ doesnot have “enough elements”. More formally, $\\mathbb{Q}$ as a metric spaceis not complete (http://planetmath.org/Complete). The $M_{2}$ defined above is the real interval $M_{2}^{\\prime}:=(\\sqrt{2},\\infty)\\subset\\mathbb{R}$ intersected with $\\mathbb{Q}$ . $M_{2}^{\\prime}$ as a subset of $\\mathbb{R}$ does have an infimum (namely $\\sqrt{2}$ ),but as that is not an element of $\\mathbb{Q}$ , $M_{2}$ does not have aninfimum as a subset of $\\mathbb{Q}$ . This example also makes it clear that it is important to clearly state thesuperset one is working in when using the notion of infimum or supremum. It also illustrates that the infimum is a natural generalization of theminimum of a set, as a set that does not have a minimum may still havean infimum (such as $M_{2}^{\\prime}$ ). Of course all the ideas expressed here equally apply to the supremum, as thetwo notions are completely analogous (just reverse all inequalities).
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sets that do not have an infimum

Proof.