# neighborhood and topological space

Let us say that there is Point $1$. And there are Points $2$, $3$ and $4$. ($X$ = $\{1,2,3,4\}$.)

And then, $\tau$ of topological space $X$ is defined as $\{\emptyset, \{1\}, \{1,2,3,4\}\}$.

The first question is, as $\{1,2,3,4\}$, $1$ would be able to reach $2$, which is in the set. Then, why is $\{1,2\}$ not automatically included in $\tau$?

The second question is, what would $\{1\}$ being an open set mean? $\{1\}$ would mean that it is possible to move somewhat and reach itself, as far as I know, and it does not seem to make sense.

I think I am getting some concepts wrong during the class, so can anyone help me correct misunderstanding?

Edit: OK, I now get some points. So, say there is a sequence $A$. The topological space is $X$, and $\tau = \{\emptyset,\{1\},\{1,2\},\{1,2,3,4\}\}$. If $A$ converges to $2$, it would converge to $1$ and $3$ and $4$. Then, why is $\{1,2\}$ needed after all? Does this mean that there is a sequence that converges only to $2$, and not $3$ and $4$?

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The topology is something that you define, and nothing automatically includes there. You only have to define a collection of sets which contains $\emptyset,X$ and close under finite intersection and any unions. I don't know what do you mean with "1 is able to reach 2" since there is no any dynamics so far, just a topological structure. The point is that topology is something which tells us: "what is convergence" and "what is continuity". Namely, in the case you've provided any sequence which converge to $2$ converges to $1$ and $3$ and $4$ since the only open neighborhood of $2$ is whole $X$. –  Ilya May 24 '12 at 10:20

Given a set $X$, a topology on $X$ is a collection of subsets of $X$ - say $\tau$ with the properties:

1. $\varnothing\in\tau, X\in\tau$.
2. If $A,B\in\tau$ then $A\cap B\in\tau$.
3. If $\mathcal U\subseteq\tau$ is any collection of elements from $\tau$, then $\bigcup\mathcal U=\{x\in X\mid\exists U\in\mathcal U: x\in U\}$ is also in $\tau$.

We say that sets are open if they are in the topology $\tau$, and closed if their complement is in $\tau$.

Now consider $X$ with the given $\tau=\{X,\{1\},\varnothing\}$. We can verify it is indeed a topology, it is closed under finite intersections and under unions.

This means that $\{1,2\}$ need not be in $\tau$ for it to be a topology, nor the complement of it (the set $\{3,4\}$) need to be in $\tau$. This set is neither open nor closed.

Again, the solution is to read the definitions and work with them one step after another:

$a_n$ converges to $a$ if whenever $U$ is open (i.e. in $\tau$) there is some $k\in\mathbb N$ such that for all $n>k$ we have $a_n\in U$.

Note that if a sequence converges in your $X$ to $1$, and $\{1\}\in\tau$ this means that the sequence is constant from one point onwards. If it converges to $2$ then it can be either $1$ or $2$ from some point.

In particular, the sequence $a_n =\begin{cases} 1 & n \text{ is even}\\ 2 & n\text{ is odd}\end{cases}$ converges to $2$ but not to $1$. Peculiar? Yes, but finite topologies are often peculiar.

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Some topologies indeed do not make sense. The chaotic or indiscrete topology, whose only open sets are $\emptyset$ and $X$, does not have any nontrivial open sets (by definition, duhh), so it's very hard to imagine a metric space having these properties. (in fact, a metric space whose topology is indiscrete necessarily has at most 1 element).