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I'm currently in a real analysis class right now, and we are learning about open and closed sets. We are discussing the concept of limit points, and I am having trouble visualizing them. I usually picture a circle with a closed, solid outline and that the limit points are contained on that outline. If a set is closed, it contains all of its limit points. I know open sets can have limit points as well, but I always imagine open sets as a circle with a dashed outline. Could someone perhaps draw me a visual representation of a limit point in relation to its $$ \epsilon- neighborhood? $$ Thank you!

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  • $\begingroup$ Okay, if you image the universe being R^2 (which is not always a good idea). There are three types of limit points. There are points that are "completely" in the set so that there is always a circle around it that's complete in the set. There are limit points on "the edge of set" so that every circle around the set has to intersect and have many points in the set. And limit points "just by the rim" these points aren't in the set but are so close that every circle intersects the set. You MUST remember closed and open are NOT opposites and "interior" and "limits" aren't either. $\endgroup$ – fleablood May 28 '16 at 21:57
  • $\begingroup$ You can have a "foggy" misty set like Q in R. Every real number, rational or not, is "right up next to" a point of Q so every point is a limit point. THis misty set is not closed because the irrational limit points are not in it. This misty set is not open either because none of it's points are "entirely" inside it so that there is a circle completely in the set. (So there are no interior points) $\endgroup$ – fleablood May 28 '16 at 22:03
  • $\begingroup$ You can have a solid set like a disk. Every point "inside" the disk is both a limit point and an interior point as there are always circles around the point that are enitrely inside the set. The edge points are not interior points but are limit points as every circle intersects both points in the set and not in the set. So the disk is open if none the edge pieces are in the set (so only the interior points are in the set). The disk is closed if all the edge peices are in the set (so all the limit points are in the set). $\endgroup$ – fleablood May 28 '16 at 22:06
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Imagine any neighborhood of a point $x$. If every neighborhood intersect a set $A$ in a different point than $x$ then $x$ is a limit point of $A$.

To simplify: in the standard topology of $\Bbb R$ an $\epsilon$-neighborhood is any open interval of the kind $(x-\epsilon,x+\epsilon)$ for any $\epsilon>0$ (the $\epsilon$-neighborhood is a basis for all neighborhood on the standard topology in $\Bbb R$, so any result that holds for $\epsilon$-neighborhoods holds too for any other kind of neighborhood).

Then if $\forall\epsilon>0, (x-\epsilon,x+\epsilon)\cap A\neq \{x\}\text{ or }\emptyset$ then $x$ is a limit point of $A$. The thing to visualize is the concept of $\epsilon$-neighborhood intersecting $A$ in a different point than $x$, this is the image for this context.

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You can imagine that if $x$ is a limit point, then don't matter how close you look, you ever found another point of the set with him.

Or you just can imagine that it is not a isolated point, that is more easy.

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This: $$( \bullet ) \hspace{-16px}\color{red}{\left(\begin{matrix}\quad\quad \\ \quad\end{matrix}\right)} \quad\quad\text{ bigger $\epsilon$ }$$ where $\color{red}{\left(\begin{matrix}\quad\quad \\ \quad\quad\end{matrix}\right)}$ is the set you want a limit point of and $( \bullet )$ is an epsilon neighborhood around the point $\bullet$.

Making the $\epsilon$-neighborhod smaller, the interval enclosed in the black parentheses will still intersect the interval with the big red parentheses.

$$( \hspace{-4px} \bullet \hspace{-4px} ) \hspace{-11px}\color{red}{\left(\begin{matrix}\quad\quad \\ \quad\quad\end{matrix}\right)} \quad\quad\text{ smaller $\epsilon$ }$$

Hopefully the diagrams render the same for all viewers.

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    $\begingroup$ But I think it's important to realize this is also a limit point: $\hspace{-16px}\color{red}{\left(\begin{matrix}\quad\quad \color{black}{( \bullet )} \\ \quad\end{matrix}\right)} \quad\quad\text{ }$ (it's bot an interior point AND a limit point.) $\endgroup$ – fleablood May 28 '16 at 22:10
  • $\begingroup$ Yes! I knew it but neg-, lected to include it so, I'm glad you drew it. $\endgroup$ – jdods May 28 '16 at 22:48
  • $\begingroup$ I think my confusion when I was a student was thinking limit points (which if all are in S means S is closed) and interior points (which if all points of S are) were different or opposite ideas somehow when in actuality they are not at all. $\endgroup$ – fleablood May 28 '16 at 23:58
  • $\begingroup$ The interior and the boundary are both always subsets of the set of limit points, yes? Regardless of choice of metric, topology, etc. $\endgroup$ – jdods May 29 '16 at 0:25

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