# 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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The point of topology, seen as a generalization of metric spaces, is that you forget about all of these dynamics. The open sets are exactly what you define as open sets, as long as your topology satisfies the usual criteria.

If you want to talk about "reaching" points, then you are probably referring to paths, path connectedness and connectedness. These are not defined using metrics, but using open sets.

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).

Perhaps you should look into the definition of "Hausdorff spaces", which are topologies that are somewhat like metric spaces, but nonetheless /can/ be very unnatural.

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