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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?

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    No. Limit points can also be interior points. –  May 20 '15 at 02:55
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    @Bye_World: That answers my question:) –  May 20 '15 at 02:57
  • @Bye_World: Just can you tell me why is boundary point a limit point? Does its neighbourhood contains infinitely many elements of a sequence? –  May 20 '15 at 03:00
  • Give me a minute to type up an answer. –  May 20 '15 at 03:04
  • According to Wikipedia, a limit point and an accumulation point are synonymous. A boundary point is different. – clay Oct 10 '18 at 22:59

3 Answers3

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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 $S'$.

  • 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 $T$. – chharvey May 20 '15 at 13:15
  • 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. – chharvey May 20 '15 at 13:39
  • My apologies. I thought $X$ was the set in question, not the entire topological space. – chharvey May 20 '15 at 14:03
  • @chharvey No worries. :) –  May 20 '15 at 14:11
  • @Bye_World. Can you explain why for the set $S={0}$ in $\Bbb R$ , $0$ is not the limit point? – CKM Jun 13 '16 at 08:29
  • @chandresh There is no neighborhood of $0$ in $\Bbb R$ which contains a point in ${0}$ different than $0$ (because there is no point in ${0}$ different than $0$). You can think of a limit point as a point where no matter how small of a neighborhood around that point you consider, there will always be some other point of your subset in it. But there can't be another point in your subset if your subset is a singleton. –  Jun 13 '16 at 11:59
  • As per your definition, here is a contradiction to your above argument, Suppose I take $x=1$ in $\Bbb R$, there is a nbd of $x$ that contains an element of $S$ ($0$ in this case). I am not able to see why this is not true. Can you clarify pl? – CKM Jun 13 '16 at 15:32
  • The definition of a limit point $x$ of a set $S$ is basically this, there exist points arbitrarily close to $x$ in $S$ that are different than $x$ itself. Clearly there are not points arbitrarily close to $0$ in ${0}$, because $0$ is the only element of ${0}$. –  Jun 13 '16 at 15:38
  • @chandresh In your example, every neighborhood of $1$ would have to contain a point in $S$. But $(\frac 12, \frac 32)$ is a neighborhood of $1$ that doesn't contain a point in $S$. Thus $1$ is not a limit point of $S$. –  Jun 13 '16 at 15:39
  • @Bye_World. Thanks. I was thinking if I can find at least one neighborhood of 1 that contains points in $S$. You clarified it by stressing $\mathbf{every}$. – CKM Jun 14 '16 at 05:23
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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.

vadim123
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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 !!!

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