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Let $X=[0,1)$ with the metric $d(x,y)=|x-y|$, and $Y=\mathbb{R}^2$ with the Euclidean metric. Define the mapping $f:X\rightarrow{Y}$ by

$f(t)=(\cos(2 \pi t + \frac{\pi}{2}), \sin(2 \pi t + \frac{\pi}{2})) \forall t\in{X}$.

Let $A=[1/2,1)$. Show that $A$ is closed in $X$ and $f(A)$ is not closed in $Y$.

Obviously $A$ is closed in $X$, but I don't know how to show that mathematically. Should I somehow show the complement of A is open?

For $f(A)$ not being closed in $Y$, I plugged in the values $\frac{1}{2}$ and $1$ into $f(A)$ and found the range to be $(-1,1)$.

Not sure if I'm on the right track... any hints and/or direction would be greatly appreciated!

Michael Hardy
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mmh0015
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  • I don't think A is closed. $\bar A = [\frac12 , 1] \neq [\frac12, 1) = A$ – Bman72 Mar 12 '14 at 21:40
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    In $X$, I think $A$ is closed since $\overline{A}^X = [\frac{1}{2},1) = A$. – Guest Mar 12 '14 at 21:49
  • So basically we have $\bar{A}=[\frac{1}{2},1)\subseteq{[0,1)}=X$. Thus, since the closure of $A$ is in $X$, then $A$ is closed in $X$? – mmh0015 Mar 12 '14 at 21:54
  • If your working space is $X$, then the closure of $A$ will always be a subset of $X$. When you want to show that $A$ is closed in $X$, look at the results you know that give sufficient conditions for $A$ to be closed. It is established in any topology course (or in any course discussing the topology of metric spaces) that $A$ is closed in $X$ if, and only if, the closure of $A$ in $X$ equals $A$. In this case it is clear that the closure of $A$ in $X$ is $[1/2, 1)$, which equals $A$. Hence $A$ is closed in $X$. – Guest Mar 12 '14 at 22:17
  • As for the range of $f$, what do you mean by $(-1,1)$ ? $f$ has values in $\mathbb{R}^2$. Unless I'm mistaken, the range of $f$ is the half-unit circle lying in the first and fourth quadrants, without the point $(0,1)$. But the point $(0,1)$ is obviously a limit point of $f(A)$ that is not in $f(A)$. Hence $\overline{f(A)} \neq f(A)$ and $f(A)$ is not closed, by the same characterization I already mentioned in my last comment. – Guest Mar 12 '14 at 22:22
  • Thank you, $f$ being in only the first and fourth quadrants makes sense now; however, I'm still a little confused as to how you know the point $(0,1)$ is not included. – mmh0015 Mar 12 '14 at 22:46
  • @user121087 The point $(0,1)$ is not included in $f(A)$ because $1 \notin A$. – Guest Mar 12 '14 at 23:03
  • Oh, of course! Thank you so much. – mmh0015 Mar 12 '14 at 23:16

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