Is the Set of Integer Outputs of sin(x) Sequentially Compact in ℝ?

[MidgetDwarf]

Hi. Someone showed me a problem today regarding sequentially compact sets in ℝ.

Ie., is the set of the image of sin(x) and x is an integer greater than one, sequentially compact? Yes or no.

What is obvious is that we know that this set is a subset of [-1,1], which is bounded. So therefore the set in question must be bounded. However, when thinking about whether the set is closed, is where we ran into issues.

My idea was to use the fact that a set is closed iff it contains all of its boundary points. We know that since, the argument of the sin function must be an integer. Then the points in this set are 'discrete', and so every point in this set is actually a boundary point. Therefore the set is closed, and consequently compact.

But this seems a little off. So I suggested to argue that the complement of the set was open., and see what happens. but there another issue here.

I think an easier approach, to argue the correct way, is to just graph the demon, and look at its behavior. The problem is that I tried graphing it into matlab, but I am unsure of how to do this by only showing the integer outputs. Can someone share the graph.

 

[Infrared]

It is not compact because $1$ is not in this set (since $sin(x)=1$ only when $x=\pi/2+2n\pi$ for integer $n$ and all of these values are irrational) but it is a limit point of this set. In general, if $a/b$ is irrational, then the multiples of $a$ are dense in $[0,b]$ mod 1.

We know that since, the argument of the sin function must be an integer. Then the points in this set are 'discrete', and so every point in this set is actually a boundary point. Therefore the set is closed, and consequently compact.

The image of a discrete set under a continuous map need not be discrete.

 

[WWGD]

I guess you can also argue it's not closed , since it's dense in [-1,1] ( I believe by equidistribution theorem; if x is Irratiinal, then for n integer, {nx ( mod 1)} is dense in [0,1]), so that its closure would equal the whole interval. Thus, not being closed and bounded, it's not compact on the Reals, by Heine-Borel.

 

[WWGD]

Iirc, compactness and sequential compactness are equivalent on metric spaces, here the Reals.