A limit point is just a accumulation point whose neighbourhood contains infinitely many elements of the sequence.
Is there any difference between boundary point & limit point? I've read in another question here that all boudary points are limit points, but is the converse true?
$\endgroup$ 53 Answers
$\begingroup$Definition of Limit Point: "Let $S$ be a subset of a topological space $X$. A point $x$ in $X$ is a limit point of $S$ if every neighbourhood of $x$ contains at least one point of $S$ different from $x$ itself."
~from Wikipedia
Definition of Boundary: "Let $S$ be a subset of a topological space $X$. The boundary of $S$ is the set of points $p$ of $X$ such that every neighborhood of $p$ contains at least one point of $S$ and at least one point not of $S$."
~from Wikipedia
So deleted neighborhoods of limit points must contain at least one point in $S$. But (not necessarily deleted) neighborhoods of boundary points must contain at least one point in $S$ AND one point not in $S$.
So they are not the same.
Consider the set $S=\{0\}$ in $\Bbb R$ with the usual topology. $0$ is a boundary point but NOT a limit point of $S$.
Consider the set $S'=[0,1]$ in $\Bbb R$ with the usual topology. $0.5$ is a limit point but NOT a boundary point of $S'$.
$\endgroup$ 10 $\begingroup$Consider the interval $[0,1]$. Each element of it is a limit point, i.e. $\alpha$ is a limit of the sequence $n_1=\alpha, n_2=\alpha, \ldots$. Only $0,1$ are boundary points.
$\endgroup$ $\begingroup$Well, as someone has figured it out by supplying the definitions of limit point and boundary point. Now if we just head toward the general set topological approach we will find that , if $\Bbb{S}$ ${\subset}$ of $\Bbb{R}$ , and if $\Bbb{X}$ be the boundary then $\Bbb{X}$=cl(S)~int ( S) . So if p is a boundary point, then p will be in $\Bbb{X}$ . And we call $\Bbb{S}$ a closed set if it contains all it's boundary points. Now as we also know it's equivalent definition that s will be a closed set if it contains all it limit point.
But that doesn't not imply that a limit point is a boundary point as a limit point can also be a interior point . Let's check the proof.
Let $\Bbb{S}$ is our set of which l is a int point . Then for `$\epsilon$>0 , N(l, $\epsilon$ ) contained in l . Now we will try to prove it contrapositively . Let l is not an int point . Then N(l, $\epsilon$ ) is not contained in $\Bbb{S}$ . Now let, €>0 then either €< $\epsilon$ or €≥ $\epsilon$
When, €< $\epsilon$ as N(l, $\epsilon$ ) is not contained in $\Bbb{S}$ , so N(l, €) is not also contained in s . It suggests that, N'(l,€) ${\cap}$ $\Bbb{S}$ = $\phi$So, l is not a limit point of $\Bbb{S}$
When, €≥ $\epsilon$ , N(l, $\epsilon$ ) is contained in N(l, €). So from here also it can be shown that , $\Bbb{S}$ ${\cap}$ N'(l,€) is $\phi$ . So l is not a limit point of $\Bbb{S}$ .
So if l is not an int point of $\Bbb{S}$ , it's not an limit point of $\Bbb{S}$ . It implies that if l is an limit point of $\Bbb{S}$ , it's an interior point of $\Bbb{S}$ .
Now, there are also some cases where the above assertion fails. So l may or may not belongs to cl ( $\Bbb{S}$ ) ~ int ( $\Bbb{S}$ )
And the whole discussion tells us that a limit point can be a boundary point but that doesn't mean every limit point is a boundary point. And that's it !!!
$\endgroup$ 3
