alex2100x
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I keep coming along this terms in a few proofs I have been reading but I don't know what it means.

E.g. \displaystyle R=\left\{\begin{matrix}sup|x|: x\in S \\ \infty

\end{matrix}\right.

Could some one explain what the inf and sup are, I haven't covered it in lectures which seems weird but I want to understand what it means so I thought I'd ask here to try and get a straightforward explanation.

Thanks.
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james22
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The supremum is the smallest number, x, such that for every s in S, s<=x. It is the smallest upper bound of the set (provided an upper bound exists).

For example the supremum of (a,b) is b. The supremum of [a,b] is b, the supremum of the natural numbers does not exist.
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atsruser
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(Original post by alex2100x)
Could some one explain what the inf and sup are, I haven't covered it in lectures which seems weird but I want to understand what it means so I thought I'd ask here to try and get a straightforward explanation.
A supremum is the generalisation of the idea of a maximum element of a set. Consider the set A = [0,1] \subset \mathbb{R}

Then 1 is the maximum of A since any element is less than or equal to 1, and any number greater than 1 isn't in the set. Note that 1 is in the set.

Now consider the set B = [0,1) \subset \mathbb{R}.

Then 1 isn't the maximum of B, and in fact the set has no maximum element, since for any x \in B, x &lt; 1 but we can also find some y \in B, x &lt; y &lt; 1 i.e. given some x \in B close to 1, we can always find another element y \in B which is a bit closer to 1. Note that 1 is not in the set, so it cannot be a maximum element.

However, even if 1 isn't the maximum, it still plays a significant role in determining the "top" of the set since if we pick any number at all smaller than 1 (e.g. m = 0.9999999999 <a googleplex 9s omitted> 9), we can find some x \in B, m &lt; x &lt; 1.

That says that 1 is an upper bound for B (or to put it another way, that B is bounded above)

2, 3, 100, etc are also upper bounds for B, but 1 is a special upper bound, because it is the smallest one; it is the least upper bound for B, since as I noted above, any number smaller than 1 *isn't* an upper bound.

Another name for the least upper bound of a set is the supremum of the set. So \sup B = 1.

All subsets of \mathbb{R} that are bounded above have a supremum that is also in \mathbb{R}. This is called the least upper bound property, and it says, more or less, that \mathbb{R} is complete, that there aren't any missing points in the number line without a number to fill them.

Some sets don't have a supremum though. Consider A \subset \mathbb{Q}, A = \{ x \in \mathbb{Q} : x^2 &lt; 2 \}. Now A is bounded above (10 is an upper bound for example) but there is no least upper bound s \in \mathbb{Q}. The obvious candidate is s = \sqrt{2}, but \sqrt{2} \notin \mathbb{Q}, so there is no supremum.

That tells us that \mathbb{Q} is not complete, that it doesn't have the least upper bound property, that there *are* missing gaps in the number line of \mathbb{Q}.

If a supremum is also a member of a set, then it is also called the maximum of the set. So with the definition of A above, \sup A = \max A = 1

And the infimum is the same, except it generalises the role of minimum.
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TeeEm
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(Original post by atsruser)
A supremum is the generalisation of the idea of a maximum element of a set. Consider the set A = [0,1] \subset \mathbb{R}

Then 1 is the maximum of A since any element is less than or equal to 1, and any number greater than 1 isn't in the set. Note that 1 is in the set.

Now consider the set B = [0,1) \subset \mathbb{R}.

Then 1 isn't the maximum of B since for any x \in B, x &lt; 1 but we can also find some y \in B, x &lt; y &lt; 1 i.e. given some x \in B close to 1, we can always find another element y \in B which is a bit closer to 1. Note that 1 is not in the set.

However, even if 1 isn't the maximum, it still plays a significant role in determining the "top" of the set since if we pick any number at all smaller than 1 (e.g. m = 0.9999999999 <a googleplex 9s omitted> 9), we can find some x \in B, m &lt; x &lt; 1.

That says that 1 is an upper bound for B (or to put it another way, that B is bounded above)

2, 3, 100, etc are also upper bounds for B, but 1 is a special upper bound, because it is the smallest one; it is the least upper bound for B, since as I noted above, any number smaller than 1 *isn't* an upper bound.

Another name for the least upper bound of a set is the supremum of the set. So \sup B = 1.

All subsets of \mathbb{R} that are bounded above have a supremum that is also in \mathbb{R}. This is called the least upper bound property, and it says, more or less, that \mathbb{R} is complete, that there aren't any missing points in the number line without a number to fill them.

Some sets don't have a supremum though. Consider A \subset \mathbb{Q}, A = \{ x \in \mathbb{Q} : x^2 &lt; 2 \}. Now A is bounded above (10 is an upper bound for example) but there is no least upper bound s \in \mathbb{Q}. The obvious candidate is s = \sqrt{2}, but \sqrt{2} \notin \mathbb{Q}, so there is no supremum.

That tells us that \mathbb{Q} is not complete, that it doesn't have the least upper bound property, that there *are* missing gaps in the number line of \mathbb{Q}.

If a supremum is also a member of a set, then it is also called the maximum of the set. So with the definition of A above, \sup A = \max A = 1

And the infimum is the same, except it generalises the role of minimum.

must definitely rep this effort and time ...
(unless you are a program)
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Zacken
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(Original post by atsruser)
<snip>
This is a brilliant explanation.
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user6
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(Original post by alex2100x)
I keep coming along this terms in a few proofs I have been reading but I don't know what it means.

E.g. \displaystyle R=\left\{\begin{matrix}sup|x|: x\in S \\ \infty

\end{matrix}\right.

Could some one explain what the inf and sup are, I haven't covered it in lectures which seems weird but I want to understand what it means so I thought I'd ask here to try and get a straightforward explanation.

Thanks.
(Original post by james22)
The supremum is the smallest number, x, such that for every s in S, s<=x. It is the smallest upper bound of the set (provided an upper bound exists).

For example the supremum of (a,b) is b. The supremum of [a,b] is b, the supremum of the natural numbers does not exist.
(Original post by atsruser)
A supremum is the generalisation of the idea of a maximum element of a set. Consider the set A = [0,1] \subset \mathbb{R}

Then 1 is the maximum of A since any element is less than or equal to 1, and any number greater than 1 isn't in the set. Note that 1 is in the set.

Now consider the set B = [0,1) \subset \mathbb{R}.

Then 1 isn't the maximum of B since for any x \in B, x &lt; 1 but we can also find some y \in B, x &lt; y &lt; 1 i.e. given some x \in B close to 1, we can always find another element y \in B which is a bit closer to 1. Note that 1 is not in the set.

However, even if 1 isn't the maximum, it still plays a significant role in determining the "top" of the set since if we pick any number at all smaller than 1 (e.g. m = 0.9999999999 <a googleplex 9s omitted> 9), we can find some x \in B, m &lt; x &lt; 1.

That says that 1 is an upper bound for B (or to put it another way, that B is bounded above)

2, 3, 100, etc are also upper bounds for B, but 1 is a special upper bound, because it is the smallest one; it is the least upper bound for B, since as I noted above, any number smaller than 1 *isn't* an upper bound.

Another name for the least upper bound of a set is the supremum of the set. So \sup B = 1.

All subsets of \mathbb{R} that are bounded above have a supremum that is also in \mathbb{R}. This is called the least upper bound property, and it says, more or less, that \mathbb{R} is complete, that there aren't any missing points in the number line without a number to fill them.

Some sets don't have a supremum though. Consider A \subset \mathbb{Q}, A = \{ x \in \mathbb{Q} : x^2 &lt; 2 \}. Now A is bounded above (10 is an upper bound for example) but there is no least upper bound s \in \mathbb{Q}. The obvious candidate is s = \sqrt{2}, but \sqrt{2} \notin \mathbb{Q}, so there is no supremum.

That tells us that \mathbb{Q} is not complete, that it doesn't have the least upper bound property, that there *are* missing gaps in the number line of \mathbb{Q}.

If a supremum is also a member of a set, then it is also called the maximum of the set. So with the definition of A above, \sup A = \max A = 1

And the infimum is the same, except it generalises the role of minimum.
this is easy and i have just done gcse maths
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