Prove that $\sup \{ \varepsilon x:\, x\in A\}=\varepsilon \sup A$

Question

Since this is an introductory course to real analysis I am looking for the most simple and direct proof. Based mostly on definitions.

Let $\varepsilon$ be a positive real number. If $A$ is a non-empty bounded subset of $\mathbb{R}$ and $B = \{\varepsilon x : x \in A\}$ then $\sup(B) = \varepsilon \sup(A)$.

Since $A$ is a non-empty subset if follows that $\sup(A)$ exists as a finite supremum. Which by definition means that $x \leq \sup(A)$ for all $x\in A$ and by the second multiplicative property since $\varepsilon$ is positive it follows that $\varepsilon x \leq \varepsilon \sup(A)$ which means that $\sup(B) \leq \varepsilon \sup(A)$.

The hard part is to show that $\sup(B) \geq \varepsilon \sup(A)$ I can't think of a way to go about that.

So maybe a one shot deal... it's true always that $$\varepsilon x = \varepsilon x \space \space \forall x \in A, \space \varepsilon \gt 0 \in \mathbb{R}$$ It's also true by algebraic manipulation and associativity that $(\varepsilon x )= ((\varepsilon)(x))$ by associativity. So it follows that $\sup(B) = ((\varepsilon) (\sup(A))$

To me I don't see any issue with showing $\varepsilon x = \varepsilon x$ can lead to the conclusion of the statement.

Answer 1

$\epsilon >0.$

Show that $\epsilon \sup A \le \sup B;$

1)$B \not =\emptyset$ , bounded;

$ \sup B$ exists.

2) $\epsilon x \le \sup B$, $x \in A$.

$x \le (1/\epsilon) \sup B$, $x \in A$.

$(1/\epsilon) \sup B$ is an upper bound for $x$.

Hence $\sup A \le (1/\epsilon) \sup B$

Answer 2

In order to avoid heavy notation, let $s=\sup A$, which exists because $A$ is bounded.

Consider $b\in B$. Then, by definition, there exists $a\in A$ with $b=\epsilon a$. In particular $a\le s$, and so $b=\epsilon a\le \epsilon s$. It follows that $\epsilon s$ is an upper bound of $B$.

Let $u$ be an upper bound of $B$. Then $u/\epsilon$ is an upper bound of $A$ (prove it) and so $u/\epsilon\ge s$. Therefore $u\ge \epsilon s$.

Answer 3

I think, it is more important here to build intuition first as to what is being said, and why it should be true. Thy key is to understand the geometry of $\{ \epsilon x \; : \; x \in A\}$. For brevity, denote the latter set by $\epsilon A$.

So, consider first the special case when $A$ is a closed interval: $A = [a_{1}, a_{2}]$. What is $\epsilon A$? It is the interval $$ [\epsilon a_{1}, \epsilon a_{2}]. $$

Generally, if $A$ not necessarily an interval, but is just known to be nonempty and bounded, and $a_{1}, a_{2}$ are its elements with $a_{1} \leq a_{2}$, then multiplication by a positive $\epsilon$ preserves the latter inequality: $$ \epsilon a_{1} \leq \epsilon a_{2}. $$

Thus, for a general (nonempty, bounded) $A$ and a positive $\epsilon$, the set $\epsilon A$ is the one that results from a rescaling of $A$ that preserves the ordering.

This boldfaced property of the rescaling directly implies that the rescaling also preserves the greatest upper bound.

Answer 4

If $S$ is a subset of $\mathbb{R}$ with a minimum, $m$, then $\epsilon m \leqslant \epsilon x$ for all $x \in S$, i.e. the set $\epsilon S$ has the minimum $\epsilon m$.

If $S$ is the set of all upper bounds of $A$, then $\epsilon S$ is the set of all upper bounds of $\epsilon A$, and the result follows.

Answer 5

So maybe a one shot deal... it's true always that εx=εx ∀x∈A, ε>0∈R

That's SOOOO close.

Note: that since $\epsilon > 0$ then we will have $c < d\iff c\epsilon < d\epsilon$ and that for $a = \frac b\epsilon \in A \iff b = a\epsilon \in B$.

And using these facts it is just a matter of pointing out the facts (and just the facts, ma'am)..

1) $A$ is non-empty if and only if $B = \{\epsilon a| a \in A\}$ is non-empty. And $A$ is non-empty, so $B$ is non empty.

2) $A$ is bounded above by some $C$ if and only if $B$ is bounded above by some $\epsilon C$.

If we want a bit more detail $A$ bounded above by $C$ means for all $a \in A$ that $a \le C$. So for all $b \in B$ where $b = \epsilon a$ for some $a \in A$ that $b \le \epsilon C$. That is, by definition $B$ is bounded above by $\epsilon C$.

3) By least upper bound property of reals $\sup A$ exists if 1) and 2) are satisfied so $\sup B$ exists if and only $\sup A$ exists.

4) $C$ is an upper bound of $A$ if and only if $\epsilon C$ is an upper bound of $B$ because.... oh, I guess we already said that two bullet points above.

5) $y=\frac w\epsilon < \sup A$ if and only if $w = y\epsilon < \epsilon \sup A$ and if $y < \sup A$ means $y = \frac w\epsilon$ isn't an upper bound, then that would be true exactly if $w = y \epsilon < \epsilon \sup A$ would mean $w$ is not an upper bound of $B$.

So $M = \sup A$ if and only if $\epsilon M = \sup B$.

.....

It's kind of tempting to notice that since $\sup A$ is an upper bound of $A$ then $\epsilon \sup A$ is an upper bound of $B$ so $\epsilon \sup A \ge \sup B$ and then figure you must prove $\epsilon \sup A \le \sup B$.

This isn't actually nescessary and can get you off on irrelevant tangents. But you can do it:

If $\sup B < \epsilon \sup A$ then

$\frac {\sup B}\epsilon < \sup A$ so $\frac {\sup B}\epsilon$ is not an upper bound of $A$.

So there is an $a \in A$ so that $\frac {\sup B}\epsilon < a \le \sup A$.

So $\sup B < a\epsilon \le \epsilon \sup A$. But $a\epsilon \in B$ so that is a contradiction.

But really, the "one shot deal" is much cleaner and ... not as misleading.