![]() But if the edge need never be reached but can be a distant unattainable goal, we can cover the desk in a required infinite number of leaps. If one of his leaps must go to the edge, then no matter how we arrange them we can pick a finite number of leaps to do it in. In $\mathbb R$ examples: Unbounded and closedĪll of $\mathbb R$. The empty set is bounded because there is no distance between any two points. I'm not sure there is really anything more to say. Any distance between two points is finite. Well, bounded is exactly what it sounds like. So for all the points that are in the empty st (all zero of them) you can't claim you *can't draw a ball around them them that is entirely in the empty set So the empty set is open. There aren't any points in the empty set that you can't draw a ball around and be entirely in the empty set. The "empty set is closed" is a little more abstract. So there aren't any limit points that are not in the empty set. No points have balls that hit the empty set (there is nothing to hit) so there aren't any limit points of the empty set. Notice the empty set is both closed and open. It might be easier to say, there aren't any limit points that are not in $\mathbb Z$. It doesn't have any limit points so all the limit points (all zero of them) are in the set so. This actually means that $\mathbb Z$ is closed. So no point is a limit point of $\mathbb Z$. If you take any real number you can draw a small ball around it and not hit any integer. Every ball around every point "hits" $X$ so every point is a limit point of $X$ and every point is in $X$ so $X$ is closed. For every point of $X$, every ball will be entirely in $X$ (there's nowhere else to be) so $X$ is open. Notice the space $X$ is both open and closed. The only points that must hit $B$ when we draw balls around them are the points $$, and those points are all in $B$. Because the limit points $0$ and $1$ are not in the set.īut the set $B=$ is closed. If we were to list all the limit points of $A= (0,1)$ we would find they are the point $0$ and $1$ and all the points in between.Ī set is closed: All its limit points are also points in the set. Limit point: If every ball around $x$ must "hit" the set $A$ then $x$ is a limit point of $A$. We call such points "limit points" of $A$. But no ball around $0$ or $1$ can miss $A$. It will not happen for the point $-0.000001$ because we can take a tiny ball around $-0.0000001$ that completely misses $A$. And it will also happen for every ball around $0.75$ (or any point $y$ so that $0\le y \le 1$). It will also happen for every ball around $1$. That will happen for every ball around $0$. Notice every ball around $0$ will have some points that are bigger than $0$ (and smaller than $1$) and these will be in $A$. Now it is possible that there is a point $x$, (or more points) in the space $X$ that may, or may not be in $A$ and has the property, that every ball around $x$ will have some points in $A$.įor instance, take $A= (0,1)$ and $x = 0$. So $A$ is open.Īs every point will have a ball around it, that means for every point there "a little further you can go". But we can do a ball and we can do a ball around every point. In fact it must be smaller than $0.0000005$. Because $a$ is so close to zero, the radius of this tiny ball must be very small. We can draw a tiny ball around $0.000001$ so that all points in the ball are in $A$. So for example if $A = (0, 1)$ and we pick a point in $A$, say $a = 0.000001$. Set can be connected or separate or any set of points.Ī set $A$ is open: if for every point $a$ of $A$ we can draw a tiny ball around $a$ and all the points in the tiny ball will be in $A$. These balls can be as tiny as we want.Ī set is a collection of points. And for every point of $x$ we can draw little balls around this point. I hope I can give an intuitive idea of "open" and "closed" and "bounded" and although in formal, tha is not misleading, and that in conjunction with a formal definition will render the formal definition less vague. " I thought closed meant but then we have [7, infinity) as closed." "when I google these concepts it's too abstract, too vague," Sorry for two answers but in response to your comment:
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