How does one express "Consider $w$ s.t. $b^{w}<y$ then $b^{w+(1/n)} < y$ for sufficiently large $n$" in quantifiers and symbols?

Question

I was reading baby Rudin exercise 7d) and it said:

Consider $w$ s.t. $b^{w}<y$ then $b^{w+(1/n)} < y$ for sufficiently large $n$

how does one express that in symbols? In particular, I am confused where the sufficiently large part should go. On the LHS or RHS of the implication? Where does it go? It seems based on the structure of the sentence that the sufficiently large goes at the RHS since its at the end of the sentence but its actually on the LHS on the implication in reality. The reason I believe this is because one has to be in the sufficiently large regime for this to be true. Thus it has to mean:

If one is in the regime where $n$ is sufficiently large (above say, $N_0$), then if $w$ has the property $b^{w}<y$ then it follows that $b^{w+(1/n)} < y$.

So in symbols my guess is that it means:

$$ \exists N_0, n \geq N_0, w , b^{w}<y, \implies b^{w+(1/n)} < y$$

is that right? Or how does one express it properly in symbols?

Answer 1

I think that the statement in the second orange box is actually my symbol statement. But you translated it wrongly in symbols. To show this, I am showing that my symbol statement is correct, while yours is wrong.

The case $ 1\geq b > 0$ is quite meaningless (we can always take $N =1$). So that I will replace $b > 0$ with $b > 1$. Now my modified statement is

$\forall\ y > 1,\ \forall\ b > 1, \forall\ w \in \Bbb{R}, b^w < y \implies$ $\exists \ N \in \Bbb{N}$ s.t. $\forall \ n \geq N$, $b^{w + 1/n} < y$.

$\underline{\text{Pick any $y, b > 1$ and $w \in \Bbb{R}$ such that $b^w < y$.}}$

Let $\epsilon = \frac{y}{b^w}-1$. $\epsilon > 0$. Since $b^{1/n} \rightarrow 1$, $\underline{\exists \ N\in \Bbb{N}}$ s.t. $\forall \ n \geq N,\ $ $b^{1/n}-1 < \epsilon$.

$\underline{\text{Pick any $n \geq N$}}$. $b^{1/n} -1< \frac{y}{b^w}-1$. Thus, $b^{1/n+w} -b^w< y-b^w$, or $\underline{b^{1/n+w} < y}$.

The most important thing is that $w, b$ and $y$ has to be fixed before picking $N$. In other words, the choice of $N$ is a function of the three variables $b, y$ and $w$.

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Let me write your statement as

$\forall\ y > 1,\ \forall\ b > 1,$ $\exists \ N \in \Bbb{N}$ s.t. $\forall \ n \geq N$, $ \forall\ w \in \Bbb{R}, b^w < y \implies$ $b^{w + 1/n} < y$.

It is wrong since $N$ is claimed to be independent of $w$. The negation of this would be

$\exists\ y > 1,\ \exists \ b > 1,$ s.t. $\forall \ N \in \Bbb{N}$, $\exists \ n \geq N$ s.t. $\exists \ w \in \Bbb{R}$ s.t. $ b^w < y \wedge b^{w + 1/n} \geq y$.

Take $y =4$, $b =2$. Pick any $N \in \Bbb{N}$. Take $w = 2-\frac{1}{2N}$. $w < 2 \implies 2^w < 4$, but $w+\frac{1}{N} > 2 \implies 2^{w+1/N}> 4$.

A stronger statement than the negation would be

$\forall\ y > 1,\ \forall \ b > 1,$ s.t. $\forall \ N \in \Bbb{N}$, $\exists \ n \geq N$ s.t. $\exists \ w \in \Bbb{R}$ s.t. $ b^w < y \wedge b^{w + 1/n} \geq y$.

, which I think is also true. Since $w$ can always chosen to be $(\log_b y )-\frac{1}{2N}$.