How does one show that the set $E = \{ y : y > 0, y^n < x \}$ is non-empty?

Question

I was going through theorem 1.21 from Rudin (without looking at the proof) and I wanted to show that the supremum of $E = \{ y : y > 0, y^n < x \}$ must be the element that satisfies $y^n = x$. But for me to even attempt that next step, I need to show that $E$ is non-empty (then I need to show its bounded) so that I can show a supremum exists (so that I can use the least upper bound LUB property of R). However, I've been stuck for a couple of hours on this.

It feels really obvious it should exist. If people can provide hints first before telling me the answer, that would be super helpful!

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I wrote this and realized I did have a very ugly proof. Any help making it more concise would be helpful! (since its such an obvious idea that there must be a simpler proof that makes this fact look obvious...or at least I believe thats true).

I can share what I've thought (but hasn't helped me so far):

I considered first the case when $x>1$. In this case it should be obvious that some element a little smaller than 1 should be enough to show $1 - \epsilon < x^{1/n}$ (i.e. by showing instead $(1 - \epsilon)^n < x$).

Since $0< 1 - \epsilon < 1$ then raising it to the power of any natural number should decrease it size so:

$$ 0 < (1 - \epsilon)^n < 1 - \epsilon < 1 < x$$

which guarantees that $1 - \epsilon \in E$. So in this case $E$ is not empty it seems (since we choose $1 > \epsilon > 0$).

Now consider $x < 1$. In this case the easiest way to show $E$ is non-empty seems to me to show again that some "arbitrarily" small element close to 0 when raised to $n$ is smaller than $x$. So notice we have $0 < x < 1$. This is useful because we can invoke the fact that Q is dense in R by focusing on $0<x$. So between any reals there is a rational. Therefore, $\exists q \in Q$ s.t. $0<q<x$. Notice that $x<1$ so is $q$. Therefore, we have $0<q<x<1$. But raising $q$ to the power of anything decreases it (and since its not zero it can never go above it nor equal it), so we have $0<q^n<q<x$ . Thus in this case $E$ is also non-empty.

Notice that if $x=1$ then $y$ s.t. $y^n = x =1$ is just $y=1$ means we don't care about showing $E$ being empty (but it is cuz Q is dense in R and E would be bounded by 1 and 0).

If someone can help me make this proof shorter that would be useful! I feel that this fact is so obvious that it should have a proof that is much simpler and cleaner to explain (i.e. that makes this question obvious). If someone has one that would be useful!

Also, I hope my proof is not incorrect in anyway...

Answer 1

Assuming that $x>0$, note that there is some $m \in \mathbb{N}$ such that ${1 \over m} < x$. Then ${1 \over m^n} \le {1 \over m}$ and so ${1 \over m}$ is in the set.

Answer 2

If $ 1 < x $ then the point $y=1$ is in set. If $ 0< x \leq 1 $ then the point $\frac{x}{2}$ is in set.

Answer 3

Note for instance that if $a < b$ ($a,b$ any two real numbers), then their average $\frac{a+b}{2}$ lies between them: $a < \frac{a+b}{2} < b$. Indeed, it follows from adding $a$ (resp. $b$) to both sides of the original inequality to get $2a < a + b$ (resp. $a + b < 2b$) and multiplying both sides by $1/2$.

Now we can simply take the smallest number between $x$ and $1$, which we denote by $\min(x,1)$. And now it is clear that there is at least one $y$ satisfying $0 < y < \min(x,1)$, namely the average $\min(x,1)/2$. And so $0 < y^n < y < x$.