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Define set $S = \{(x,y) \in \mathbb{R}^2| x \in \mathbb{Z}, y = 0\}$, what is the closure of $S$?

Intuitively, this set has no interior since any ball around any point in the set will contain points outside of the set. Then this means every point is a boundary point.

What about closure? The closure is the set of limit points of $S$, and $x$ is a limit point if every ball contains a point $q \in S, q \neq x$. Clearly, not every ball can achieve this. Then the closure is also empty.

But the boundary = closure - interior....but RHS is empty. So big problem.

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  • $\begingroup$ in some sense the point at $\infty$ is in the closure $\endgroup$ – reuns Feb 28 '16 at 3:10
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You do not have the proper definition of the closure. The closure of a set $S$ is the smallest closed set containing $S$. In the case of metric spaces, this is equivalent to the set $S$ together with its limit points.

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The closure contains the set of limit points, however, they are not always equal. For example, if $X=(0,1)\subset \mathbb{R}$ in Eucledian topology, then the closure of $X$ is $[0,1]$, where the limit points of $X$ are $\pm 1$.

In general, (closure of $S$)=S $\cup$ (limit points of $S$)=(interior of $S$) $\cup$ (boundary of $S$).

When you say the closure is the set of limit points, you lose information. That is the source of the problem.

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  • $\begingroup$ My example was wrong. The set of limit points of $(0,1)$ is $[0,1]$ just like its closure. Please check the definitions. I was thinking in a different way. A trivial example would be $X=\{0\}$. It has no limit points, however, its closure is again $\{0\}$. $\endgroup$ – Mustafa Feb 28 '16 at 3:08
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There are two different definitions of "limit point" that you're confusing!

If you want to think of the closure of a set $A$ as the collection of all limit points, then your definition of "limit point" isn't the right one! You have to use the definition that $x$ is a limit point of $A$ so long as the ball $B(x, \varepsilon)$ has nonempty intersection with $A$ for all $\varepsilon > 0$. (Or you could use an equivalent definition, such as a limit point being the limit of a sequence contained in $A$.) Notice that, in this definition, it's ok if $B(x, \varepsilon) \cap A = \{x\}$.

Under this definition, the limit points of $\mathbb{Z}$ are simply all of $\mathbb{Z}$ itself. (And in fact, every point in a set is always a limit point of that set.) This now satisfies the fact that the boundary should be the closure minus the interior -- your "equation" is just $\mathbb{Z} = \mathbb{Z} - \varnothing$.

As a good exercise, you can try to prove that the closure of $A$ as we defined it (meaning the set of all limit points of $A$) is also the intersection of all closed sets containing $A$. (Some authors use this "intersection" as the definition of the closure of a set.)

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