8
$\begingroup$

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?

$\endgroup$
5
  • 4
    $\begingroup$ No. Limit points can also be interior points. $\endgroup$
    – user137731
    May 20 '15 at 2:55
  • $\begingroup$ @Bye_World: That answers my question:) $\endgroup$
    – user142971
    May 20 '15 at 2:57
  • $\begingroup$ @Bye_World: Just can you tell me why is boundary point a limit point? Does its neighbourhood contains infinitely many elements of a sequence? $\endgroup$
    – user142971
    May 20 '15 at 3:00
  • $\begingroup$ Give me a minute to type up an answer. $\endgroup$
    – user137731
    May 20 '15 at 3:04
  • $\begingroup$ According to Wikipedia, a limit point and an accumulation point are synonymous. A boundary point is different. $\endgroup$
    – clay
    Oct 10 '18 at 22:59
11
$\begingroup$

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
  • $\begingroup$ Let it be known that boundary points need not be members of $X$. For example, the set $T=(0,1)$ has two boundary points: $0$ and $1$, even though they are not members of $T$. $\endgroup$
    – chharvey
    May 20 '15 at 13:15
  • $\begingroup$ Yes, that's true. From a purely answer-the-OP's-question point of view, it helps to make a distinction, but from an educational point of view (for others who might later stumble upon your answer) it should be technically correct. $\endgroup$
    – chharvey
    May 20 '15 at 13:39
  • $\begingroup$ My apologies. I thought $X$ was the set in question, not the entire topological space. $\endgroup$
    – chharvey
    May 20 '15 at 14:03
  • $\begingroup$ @chharvey No worries. :) $\endgroup$
    – user137731
    May 20 '15 at 14:11
  • $\begingroup$ @Bye_World. Can you explain why for the set $S={0}$ in $\Bbb R$ , $0$ is not the limit point? $\endgroup$
    – CKM
    Jun 13 '16 at 8:29
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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
  • $\begingroup$ Does this mean that, in the standard topology, every point in an open set is a limit point? $\endgroup$ Jan 13 at 22:29
  • $\begingroup$ Yes!Ralph . In point-set topology, every point of an open set is limit point and at the end of the day , what really makes the difference between an open and close set is exclusion and inclusion of boundary, respectively. $\endgroup$
    – Gogole Pi
    Jan 14 at 3:53
  • $\begingroup$ Google Pi, thank you. $\endgroup$ Jan 15 at 4:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy