Question about limits involving 1/x

[docnet]

Homework Statement

$\frac{1}{x}$

Relevant Equations

$\frac{1}{x}$

For values of $x$ such that $x>0$, is $\frac{1}{x}$ bound above?

My reaction is that if $x>0$ then $\frac{1}{x}$ is defined because $x\neq 0$. But, it is not bound above because $x$ can be taken arbitrarily close to $0$ and $\displaystyle{\lim_{x \to 0}} \frac{1}{x} = \infty$

 

[fresh_42]

Homework Statement:: $\frac{1}{x}$
Relevant Equations:: $\frac{1}{x}$

For values of $x$ such that $x>0$, is $\frac{1}{x}$ bound above?

My reaction is that if $x>0$ then $\frac{1}{x}$ is defined because $x\neq 0$. But, it is not bound above because $x$ can be taken arbitrarily close to $0$ and $\[ \lim_{x \to 0} \frac{1}{x} = \infty \]$

Yes. And what is the question?

You can formally say: Given $M > 0,$ then for all $0< x < 1/M$ we get $\dfrac{1}{x}>M$. This means that we can find a number $1/x$ which is greater than any boundary $M.$

 

[Prof B]

Homework Statement:: $\frac{1}{x}$
Relevant Equations:: $\frac{1}{x}$

For values of $x$ such that $x>0$, is $\frac{1}{x}$ bound above?

My reaction is that if $x>0$ then $\frac{1}{x}$ is defined because $x\neq 0$. But, it is not bound above because $x$ can be taken arbitrarily close to $0$ and $\displaystyle{\lim_{x \to 0}} \frac{1}{x} = \infty$

$\displaystyle{\lim_{x \to 0}} \frac{1}{x}$ is undefined
$\displaystyle{\lim_{x \to 0^+}} \frac{1}{x} = \infty$
Follow your teacher's instructions, but for questions like this I think you are better off using the actual definition of unbounded.

 

[docnet]

Yes. And what is the question?

You can formally say: Given $M > 0,$ then for all $0< x < 1/M$ we get $\dfrac{1}{x}>M$. This means that we can find a number $1/x$ which is greater than any boundary $M.$

Kind of related to limits, how does one numerically/symbolicaly write "the smallest element greater than x"? Since the real numbers exist in a continuity, the intuition seems like there should be one single irrational number that is "touching x" like in the case of the integers, "x+1 is touching x"

 

[PeroK]

Kind of related to limits, how does one numerically/symbolicaly write "the smallest element greater than x"? Since the real numbers exist in a continuity, the intuition seems like there should be one single irrational number that is "touching x" like in the case of the integers, "x+1 is touching x"

There is no such number. The set $\{y \in \mathbb R: y > x \}$ has no least member.

In other words, that is a set without a minimum. The number $x$, however, is the infimum of the set. The infimum is also know as the greatest lower bound. Compare with supremum, which is the least upper bound.

These are absolutely critical concepts in standard real analysis.

 

[docnet]

There is no such number. The set $\{y \in \mathbb R: y > x \}$ has no least member.

These are absolutely critical concepts in standard real analysis.

I understand why.. given a $y>x$, you can always write a smaller number $\frac{y-x}{2}+x > x$. This has been bothering me because math seems to be broken somehow. It's weird that a set of numbers has a lower bound, but not a minimum element.

 

[PeroK]

I understand why.. given a $y>x$, you can always write a smaller number $\frac{y-x}{2}+x > x$. This has been bothering me because math seems to be broken somehow. It's weird that a set of numbers has a lower bound, but not a minimum element.

Even the rationals have that property. There is no least rational greater than $0$, for example.

That one's easy, because $0 < \frac m {2n} < \frac m n$.

 

[fresh_42]

There is no such number. The set $\{y \in \mathbb R: y > x \}$ has no least member.

In other words, that is a set without a minimum. The number $x$, however, is the infimum of the set. The infimum is also know as the greatest lower bound. Compare with supremum, which is the least upper bound.

These are absolutely critical concepts in standard real analysis.

... and there is the axiom of choice. I even bought a book because it contains the equivalence proofs of the various versions. And it turned out to be a highly technical analysis book (Hewitt, Stromberg).

 

[Mark44]

This has been bothering me because math seems to be broken somehow. It's weird that a set of numbers has a lower bound, but not a minimum element.

IMO, it's not math that's broken, but rather your concept of it is flawed. What you're describing is the difference between discrete sets such as the integers, which have uniform separations between adjacent members, versus sets such as the real numbers.

 

[nuuskur]

This has been bothering me because math seems to be broken somehow. It's weird that a set of numbers has a lower bound, but not a minimum element.

I might find some things in math strange, but it's not sufficient to blame math when it doesn't agree with intuition. $\mathbb R$ is not well ordered with respect to the natural order.

 

[Delta2]

It's weird that a set of numbers has a lower bound, but not a minimum element.

Yes it is kind of weird, but that's why one of the axioms on the set of real numbers is that every bounded set has its infimum and supremum, but not necessarily minimum or maximum. So minimum or maximum are the infimum (or supremum) that also belong to the set.

For example if we consider the set $$Y(x)=\{y \in \mathbb{R}: y>x\}$$ it is infimum of Y(x)=x but the minimum doesn't exist because $x$ doesn't belong in $Y(x)$ as x>x doesn't hold. If we change its definition to $$Y(x)=\{y \in \mathbb{R}: y\geq x\}$$ then the infimum and minimum exist and they are equal to $x$.

And no, it doesn't matter if the Y set is countable or uncountable, or x is rational or irrational. To see this consider the set $$Y=\{\frac{1}{n} ,n\in \mathbb{N}\}$$. The set Y is countable, n is integer, 1/n is rational but it doesn't have minimum either, the infimum is 0, but there is no $n\in\mathbb{N}$ for which $$\frac{1}{n}=0$$.