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I am a beginner in the concept of general Topology. I have some confusion about the utilization of neighborhood and open neighborhood.

$\mathbf{Definition}$: Let $(X,\mathcal{T})$ be a topological space and $x\in X$. $N\subset X$ is called neighborhood of $x$ if there exists $U\in \mathcal{T}$ such that $x\in U\subset N$. It is clear that any open set $U\in \mathcal{T}$, such that $x\in U$, is a neighborhood of $x$.

Now, Let $(x_n)_{n\in\mathbb{N}}\subset X$ be a sequence and $x\in X$.

$\mathbf{Definition\ 1}$: $x$ is a limit of the sequence $(x_n)$ if for any neighborhood $N$ of $x$, there exists $n_0\in\mathbb{N}$ such that $x_n\in N$ for all $n\geq n_0$.

Some people use a different definition:

$\mathbf{Definition\ 2}$: $x$ is a limit of the sequence $(x_n)$ if for any open neighborhood $N$ of $x$, there exists $n_0\in\mathbb{N}$ such that $x_n\in N$ for all $n\geq n_0$.

The only difference is that, in the Definition 1, they consider all neighborhoods of $x$ where in the Definition 2, they only consider open neighborhoods. But is that difference not a problem?

Are those definitions equivalent? The same for the definition of Adherent point.

$\mathbf{Definition\ 1}$: $x$ is an adherent point of $A$ if for any neighborhood $N$ of $x$, $N\cap A\neq \emptyset$.

$\mathbf{Definition\ 2}$: $x$ is an adherent point of $A$ if for any open neighborhood $N$ of $x$, $N\cap A\neq \emptyset$.

I am really confused about those two definitions.

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4 Answers 4

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In both cases the definitions are equivalent. It should be clear in each case that if Definition $\mathbf{1}$ is satisfied, then so is Definition $\mathbf{2}$.

Suppose in the first case that for each open nbhd $U$ of $x$ there is an $m_U\in\Bbb N$ such that $x_n\in U$ whenever $n\ge m_U$, and let $N$ be any nbhd of $x$. By definition there is an open nbhd $U$ of $x$ such that $U\subseteq N$, and clearly $x_n\in N$ whenever $n\ge m_U$.

Finally, suppose in the second case that each open nbhd of $x$ has non-empty intersection with $A$, and let $N$ be any nbhd of $x$. As before, there is an open nbhd $U$ of $x$ such that $U\subseteq N$, and clearly $N\cap A\supseteq U\cap A\ne\varnothing$, so $N\cap A\ne\varnothing$.

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  • $\begingroup$ You made it very clear to me. Thank you so much. $\endgroup$
    – Cachorro
    Commented Jan 2, 2017 at 23:27
  • $\begingroup$ @Cachorro: You’re very welcome; I’m glad that it helped. $\endgroup$ Commented Jan 2, 2017 at 23:27
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The two definitions are equivalent. It's clear to imply the conclusion of open neighborhoods from that of neighborhoods. Now conversely, if the statement (in particular, your definitions) holds for open neighborhoods of $x,$ then for any neighborhoods, one can always find an open set containing $x,$ which is an open neighborhood of $x.$ Hence the conclusion follows.

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It should be clear that if $x$ is the limit of $(x_n)$ by Definition $1$, then Definition $2$ is satisfied as well (if it's not clear let me know and I can explain). Hence we only need to show that if Definition $2$ is satisfied, then Definition $1$ is as well:

Suppose $x$ is the limit of $(x_n)$ by Definition $2$. Let $N$ be any neighborhood of $x$, so there exists some open $U$ with $x\in U\subset N$. Because $U$ is an open neighborhood of $x$, by assumption there exists $n_0\in\Bbb{N}$ such that $x_n\in U$ for $n\ge n_0$. But $U\subset N$, so $x_n\in N$ for $n\ge n_0$ as well, and Definition $1$ is indeed satisfied.

So we see the two definitions of a limit are indeed equivalent. I'd recommend you try to apply a similar argument to your second pair of definitions.


Edit: Here is the argument to show Definition $1$ implies Definition $2$:

Suppose Definition $1$ holds. Then if $U$ is any open neighborhood of $x$, because $U$ is a neighborhood Definition $1$ tells us that there is an $n_0$ such that $x_n\in U$ for $n\ge n_0$; since $U$ was arbitrary, we see Definition $2$ holds.

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  • $\begingroup$ Can you explain to me why if $x$ is a limit of $(x_n)$ according to the Definition 1 then it is a limit of $(x_n)$ according to the Definition 2, please? $\endgroup$
    – Cachorro
    Commented Jan 2, 2017 at 23:17
  • $\begingroup$ @Cachorro because if the property holds for ALL neighborhoods (Definition $1$), then it surely holds for open neighborhoods as well (Definition $2$). $\endgroup$ Commented Jan 2, 2017 at 23:18
  • $\begingroup$ I got it. Thank you so much. $\endgroup$
    – Cachorro
    Commented Jan 2, 2017 at 23:26
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These two Definitions are equivalent. To see this, consider a sequence $(x_n)_{n \in \mathbb{N}}$ with limit $x$. I start by showing $$Definition 1 \implies Definition 2$$

By definition 1 we know that for any neighborhood $N$ of $x$, $\exists \;n_0 \in \mathbb{N} \;$ s.t. $\;x_n \in N \; \forall n≥n_0$.
Now pick one such neighborhood $N$. By the definition of neighborhood, $\exists U$ in the topology s.t. $x \in U \subset N$. In particular, since $U$ is in the topology, $U$ is an open set. But $U$ is also a neighborhood itself and chosen randomly. Hence Definition 2 follows.
Remark: a faster way to see this is to realize that the set of all open neighborhoods of $x$ is a subset of the set of all neighborhoods of $x$.
Now to see $$Definition 2 \implies Definition 1$$
Take any open neighborhood $U$ of $x$. Then by definition 2 $\exists \; n_0 \in \mathbb{N} \;s.t. \; x_n \in U \; \forall n≥n_0$. Now if you consider any arbitrary neighborhood $N$ of $x$ containing $U$, i.e. $U \subset N$, you clearly also have that $x_n \in N \forall n≥n_0$, since $(x_j \in U \implies x_j \in N )\forall j \in \mathbb{N}$. Thus also definition 1 holds and we finally conclude that these definitions are equivalent.
The equivalence of the definitions about an adherent point can be shown quite analogously.

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