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Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function.

$A=\{(x,y)\in\ \mathbb{R}^2:x^2+y^5=1$}

$B=\{(x,x\sin\frac{1}{x})\in\ \mathbb{R}^2:0<x\leq1\}\cup\{(0,0)\}$

$C=\{x\in\mathbb{R}:f(x)=0\}$

(1) Which of the above set are necessarily open?

(2) Which of the above set are necessarily closed?

(3) Which of the above set are necessarily compact?

(4) Which of the above set are necessarily connected?

I think that A is closed, compact and connected. Am I right?

And I have no idea for B.

For C,

If $f(x)=\sin x$,

$C=\{n\pi:\text{ for all } \ n\in\mathbb{Z}\}$ , so it is not connected and compact.

How to check it is open or closed?

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    $\begingroup$ hint on C: the preimage of a closed set wrt a continuous function is closed. $\endgroup$
    – drhab
    Jun 9, 2017 at 7:57
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    $\begingroup$ To show that B is closed, use the same argument as in A and C by showing that $g: \mathbb{R} \rightarrow \mathbb{R}$ by $g(0) = 0, g(x) = x\sin(\frac{1}{x})$ is continuous on $[0, 1]$. It will be compact if it's closed, since it's bounded. For connectedness of A, show that the part in the right half and left half of the plane are each connected, and that they intersect at $(0,1)$. Thus their union is connected. To show that B is connected, the image of $(0,1]$ is connected (continuous image of connected is connected). Use that the closure of a connected set is connected. $\endgroup$
    – John Samples
    Jun 9, 2017 at 8:21

1 Answer 1

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C is the inverse image of a closed set {0}
by a continous function, hence closed.

Same is true of the circle A, which also
being bounded makes it compact.

B is not open. It is closed and bounded, hence compact.
The point (1,sin 1) is not an interior point. B is also compact.

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  • $\begingroup$ B is not the topologist's sine curve; here $y(t) = t \sin(\frac{1}{t})$, not $\sin(\frac{1}{t})$. It is path-connected. $\endgroup$
    – John Samples
    Jun 9, 2017 at 8:08
  • $\begingroup$ Be careful $(0,1)$ is not in the closure of $B$ since $B$ is the graph of $x\mapsto x\sin \frac1x$ and not the graph of $x\mapsto \sin\frac1x$ as you maybe thought. $\endgroup$
    – Mundron Schmidt
    Jun 9, 2017 at 8:11

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