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Let $S, T\subseteq \mathbb{R}$ be given by \begin{align*} S &= \left\{x\in \mathbb{R}:2x^2\cos\frac{1}{x} = 1\right\} \\ W &= \left\{x\in \mathbb{R}:2x^2\cos\frac{1}{x} \leq 1\right\}\cup\{0\} \end{align*}

Under the usual metric of $\mathbb{R}$: I have to check for completeness, compactness and connectedness of subset $S$ and $T$.

I am fully stucked on this problem. Even I am finding it difficult to visualize these sets. I have got this problem from one of entrance exam paper.

Thanks for any help

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For the first: setting $y=1/x$ and rearranging gives $\cos(y)=y^2/2$. You should be able to graph these two functions and work out what $S$ looks like. –  Chris Eagle Jun 3 '12 at 9:25
    
For the first, $2x^2 \cos \frac{1}{x}$ can only hit 1 when $x$ is far enough away from 0, which is the confusing part of the function. For $x$ big, each wiggle of $\cos$ will give you several preimages of 1, so $S$ is just a countable discrete set-complete, not compact, not connected. –  uncookedfalcon Jun 3 '12 at 9:28
    
@uncookedfalcon: for $x$ big, $\cos$ will stop wiggling... –  Ben Millwood Jun 3 '12 at 9:29
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@SteveD I am a member of this community for the last 27 days(27 days consecutive) and i had asked 71 questions so for. You may calculate average of number of questions asked by me in one day. I dont think asking two or three questions per day will harm this community. –  srijan Jun 3 '12 at 14:09
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@SteveD: That is more of a meta question, and sure enough it has come up: meta.math.stackexchange.com/q/2464 Supposedly there is a limit of 6 questions per day, something srijan isn't close to. $@$srijan: Do not worry, your questions are welcome. Thank you for helping to add good content to the site. –  Jonas Meyer Jun 4 '12 at 8:12

2 Answers 2

up vote 3 down vote accepted

The easy way to get started is to let $$f(x)=2x^2\cos\frac1x$$ and observe that this is a continuous function on $\Bbb R\setminus\{0\}$. $S=f^{-1}\big[\{1\}\big]$, the inverse of a closed set under a continuous function, so $S$ is closed. Similarly, $W=f^{-1}\big[(\leftarrow,1]\big]$, so it’s also the inverse image of a closed set and hence closed. It’s clear that $$\lim_{x\to\pm\infty}f(x)=\infty\;,$$ so $S$ and $W$ must be bounded and hence compact. Thus, it only remains to determine whether they are connected. This requires some closer examination of the function $f$ and the sets $S$ and $W$. I’ll do more than is actually necessary $-$ enough, in fact, to answer the question even without using the shortcut above to see that $S$ and $W$ are compact.

If we can figure out what $S$ looks like, it probably won’t be too hard to see what $W$ looks like, so let’s begin with $S$. The condition that $$2x^2\cos\frac1x=1$$ can be rewritten as $$2x^2=\sec\frac1x\;.$$ Now either make a rough sketch or use graphing software to graph $y=2x^2$ and $y=\sec\frac1x$ and see where the graphs cross. You’ll see that they appear to cross in just two places, one on each side of the $y$-axis. Let’s see if we can verify this.

First, both functions are even, so we can limit our attention to $x\ge0$: if they intersect at some positive $x$, they will also intersect at $-x$, and vice versa. Now $\left|\cos\frac1x\right|\le1$ for all $x$, so $\left|\sec\frac1x\right|\ge1$ for all $x$. Clearly $2x^2\ge0$ for all $x$, so we can only get an intersection at values of $x$ for which $\sec\frac1x\ge1$.

$2x^2\ge1$ iff $x^2\ge\frac12$ iff $|x|\ge\frac1{\sqrt2}$, so any intersection with $x\ge0$ must occur at values of $x\ge\frac1{\sqrt2}$. On the other hand, if $x\ge\frac1{\sqrt2}$, then $\frac1x\le\sqrt2<\frac{\pi}2$. Thus, as $x$ increases from $\frac1{\sqrt2}$ towards $\infty$, $2x^2$ increases monotonically from $1$ towards $\infty$, while $\frac1x$ decreases monotonically from $\sqrt2$ towards $0$, and therefore $\sec\frac1x$ decreases monotonically from $\sec\sqrt2$ towards $\sec0=1$ (where $\sec\sqrt2\approx 6.41257$). Clearly these graphs must cross exactly once for $x\ge0$, say at $x=a$, and once again at the symmetrically opposite point $x=-a$ to the left of the $y$-axis. Thus, $S$ is a $2$-point set, $\{-a,a\}$, and as such is compact, complete, and not connected.

Now $$W = \left\{x\in \mathbb{R}:2x^2\cos\frac{1}{x} \leq 1\right\}\cup\{0\}\;.$$

Certainly $2x^2\cos\frac1x\le1$ whenever $\cos\frac1x\le0$. When $\cos\frac1x>0$ you can divide by the cosine to get $$2x^2\le\sec\frac1x\;.$$ In the first part we saw that $2x^2\le1$ for $|x|\le\frac1{\sqrt2}$; thus, if $0<\frac1{\sqrt2}$, either $\sec\frac1x<0$, or $2x^2\le1\le\sec\frac1x$, and in either case $x\in W$. Moroever, we saw that for $\frac1{\sqrt2}\le x\le a$, $2x^2\le\sec\frac1x$, so again $x\in W$. On the other hand, we saw that $2x^2>\sec\frac1x$ when $x>a$, so $W\cap(a,\to)=\varnothing$. Finally, $0\in W$ by definition, on $W$ is symmetric about the origin, so $W=[-a,a]$ and is compact, complete, and connected.

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if $x\rightarrow \infty$ imply $f(x)\rightarrow \infty$ then how can we say that function is bounded? please explain me sir. –  srijan Jun 3 '12 at 12:17
    
@srijan: The function is unbounded, but the sets $S$ and $W$ are bounded: one is $[-a,a]$, and the other is $\{-a,a\}$. The fact that the function is unbounded helps to show that the sets are bounded. (It really helps to have a picture of the graph.) –  Brian M. Scott Jun 3 '12 at 12:21
    
"The fact that the function is unbounded helps to show that the sets are bounded":only this line is troubling me. –  srijan Jun 3 '12 at 12:28
    
$x\rightarrow\infty\Rightarrow f(x)\rightarrow\infty$ so $\{x, f(x)<1\}$ is bounded. It's easy to see this fact on a drawing. A "proof" by reductio ad absurdum : suppose $\{x, f(x)<1\}$ unbounded, so for all $M>0$, exists $x>M$ such as $f(x)<1$. Contradiction with $f(x)\rightarrow \infty$ when $x\rightarrow\infty$. –  Kris H. Jun 3 '12 at 16:32

They are inverse image of closed by continuous maps. So they are closed. They are bounded (if $|x|\rightarrow +\infty$, functions are too). So they are compact subsets. And so they are complete.

$S$ isn't connected because it have positive and negative elements, but haven't 0.

$W$ is connected : you can show it's a line segment, just study the function.

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I am sorry but I am not getting your answer. –  srijan Jun 3 '12 at 9:45
    
er what's different srijan? I'm getting what Kris is –  uncookedfalcon Jun 3 '12 at 9:46
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here's a relevant plot for $W$: wolframalpha.com/input/…, so the line segment thing seems legit, you can prolly spool it out via calculus –  uncookedfalcon Jun 3 '12 at 9:47
    
@uncookedfalcon Thanks for graph. It is helpful to me. –  srijan Jun 3 '12 at 12:48

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