Show that every open set A in a metric space (X,d) is the union of closed sets.
I am beyond confused for this question. I thought union of closed sets are closed
[Math] Real Analysis: open and closed sets
general-topology
Related Solutions
No, your argument is not correct. I’ll comment on it in detail.
Consider the interval $E=[0,1]$ which is closed. By definition, every point of $E$ is a limit point.
While it is true that every point of $[0,1]$ is a limit point of $[0,1]$, this is not a matter of definition: there is nothing in the definition of closed set that requires every point of a closed set to be a limit point of that set. Moreover, $[0,1]$ isn’t a subset of $\Bbb R^2$. I suspect that you’re thinking of the set $[0,1]\times\{0\}$, the set of points on the $x$-axis between $0$ and $1$ inclusive.
It is proved that all points in $R^2$ are limit points.
Limit points of what? They’re all limit points of $\Bbb R^2$, but that has nothing to do with $E$, which isn’t even a subset of $\Bbb R^2$. And if your $E$ was really supposed to be $[0,1]\times\{0\}$, then it’s not true that all points of $\Bbb R^2$ are limit points of $E$: if they were, they’d have to be in $E$, since $E$ is closed, and a closed set contains all of its limit points.
Therefore $E^c$ must contain all limit points.
Again, all limit points of what? And what does $\Bbb R^2$ have to do with $E^c$? $E^c=(\leftarrow,0)\cup(1,\to)$, a subset of $\Bbb R$, not of $\Bbb R^2$. For that matter, why are you bringing in $E^c$ at all? The question is whether every point of $E$ is a limit point of $E$, and $E^c$ isn’t going to help you to answer that question.
But $E^c$ by definition is open ($E^c=R^2\setminus E$).
See above: $E^c=\Bbb R\setminus E$, not $\Bbb R^2\setminus E$.
This is a contradiction because $E^c$ has 2 limit points,
No, every point of $(\leftarrow,0]\cup[1,\to)$ is a limit point of $E^c$; that’s far more than two limit points!
namely $E=[0,1]$.
No, the only points of $E$ that are limit points of $E^c$ are $0$ and $1$; the points of $E$ that are in $(0,1)$ are not limit points of $E^c$.
Therefore E contains points which are not limit points.
This doesn’t follow from anything that you’ve said, and in fact it contradicts your very first claim about $E$ up above. Moreover, it isn’t true, assuming that you’re talking about limit points of $E$: every point of $E$ is a limit point of $E$.
In order to answer the question, you must do one of two things: either prove that whenever $E$ is a closed subset of $\Bbb R^2$, then every point of $E$ is a limit point of $E$, or find an example of a closed subset $E$ of $\Bbb R^2$ containing a point that is not a limit point of that set $E$. You won’t be able to do the former, because it isn’t true. For the latter, just take any non-empty finite subset of $\Bbb R^2$, for instance $E=\{\langle 0,0\rangle\}$: $E$ is closed, and it has no limit points at all. In particular, $\langle 0,0\rangle$ is not a limit point of $E$, because it is not true that every open neighborhood of $\langle 0,0\rangle$ contains a point of $E$ different from $\langle 0,0\rangle$. If you want $E$ to be an infinite set, you can use $E=\big(\Bbb R\times\{0\}\big)\cup\{\langle 0,1\rangle\}$, for instance. This is a closed set, but $\langle 0,1\rangle$ is a point in $E$ that is not a limit point of $E$: if $\epsilon\le 1$, the open ball of radius $\epsilon$ centred at $\langle 0,1\rangle$ does not contain and point of $E$ besides the point $\langle 0,1\rangle$ itself. For a more extreme example, let $E=\{\langle m,n\rangle:m,n\in\Bbb Z\}$. You should try to prove that $E$ is an infinite closed set that has no limit points at all.
A closed set always contains all of its limit points. That is, suppose that $E$ is closed. If $p$ is a limit point of $E$, then $p\in E$. The point of this exercise is to show that the converse is not true: there can be a point $p\in E$ that is not a limit point of $E$.
Your approach is a good start! If you succeed in proving those two sets are open, you'll indeed be able to conclude their union is open and that their complement is closed.
Hint: To prove a set $\mathcal{O}$ is open in the ordinary Euclidean metric, you may take an arbitrary point in $\mathcal{O}$ and show that there is an open disc around it completely within $\mathcal{O}$.
An arbitrary point in the first set is an $(x,y)$ which lies below the $x$-axis. Can you find a radius around this point that stays below the $x$-axis?
Do something similar for the other set, where the point is of the form $(x,y)$ with $y>x$.
Sketching pictures and using a little simple geometry should also be helpful.
Best Answer
Finite unions of closed sets are indeed closed.
Hint: Construct an open ball around a point as a union of (increasing) closed balls around the same point. Then notice that every open set is a union of open balls.