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I stumble upon this problem which says:

The set $\{ x\in\mathbb R: x\sin x\le 1, x\cos x \le 1\}$ is contained in $\mathbb R$.Then which of the following about the set is true:

  1. a bounded closed set
  2. a bounded open set
  3. an unbounded closed set
  4. an unbounded open set.

Any kind of hints will be helpful.

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    What do you know about the definition of "closed" and "open" sets?2012-11-19
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    The inverse image of a closed set under a continuous function is closed.2012-11-19

1 Answers 1

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The inverse image of a closed set under a continuous function is closed.

For $x$ positive, just look at intervals where the sine and cosine are both negative, and those would be included, so the set is unbounded.

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    @Thomas Andrews What I know about open and closed sets is that open set does not contain any of its accumulation points whereas closed set does contain its limit points.2012-11-19
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    I am confused about the second part of the problem. Since x belongs to R and R is both open and closed set( clopen), so what should be the correct answer between (3) and (4).2012-11-19
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    The set $\{x\in\mathbb{R} \mid x\sin x\le1\}$ is the inverse-image of the closed set $(-\infty,1]$ under the continuous function $x\mapsto x\sin x$. A similar thing applies if we'd had cosine there instead of sine. Then if you want both conditions to hold, you have the intersection of those two sets, and the intersection of two (or finitely many) closed sets is closed.2012-11-19
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    *open set does not contain any of its accumulation points*... Well, no. At least, not in the usual sense of *accumulation point*.2012-11-19
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    @did sir,will you explain "not in the usual sense of accumulation point".I can not understand it.2012-11-19
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    @MichaelHardy Thank you sir.2012-11-19
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    *will you explain "not in the usual sense of accumulation point".I can not understand it*... What is your definition of *an accumulation point*?2012-11-19
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    @did sir,what i know as my graduate study book says : A real number x, which may or may not belong to S( a set of real numbers contained in R) is called an accumulation point if every nbd of x contains at least one point of S distinct from x.2012-11-19
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    Then an open set contains an awful lot of its accumulation points. For example, (0,1) is open and **every** point in (0,1) is an accumulation point of (0,1). (By the way, do you plan to apply the recommendation in my comment to your post? If you choose to ignore it, I see no reason to continue to answer your queries.)2012-11-19
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    Let A={x€R: x sinx <=1} and B={x€R; x cosx<=1}.Now if we choose x to lie in the interval (pi+2n* pi,3*pi/2+2n*pi),then we see that both A and B lie in (-∞,1] as for those values of x which lie in the aforementioned interval,both sinx and cosx will be negative.Now we see that both A and B are closed (semi-closed ). Since intersection of two closed sets is closed ,we can conclude that the set must be closed and unbounded as well.2012-11-20