Consider $\Bbb R$ with usual topology. Now consider $\Bbb N$ (set of natural numbers) with subspace topology. Let $x \in \Bbb N$ be a point in $\Bbb N$. The result says that the only possible connected set containing $x$ is {$x$} only. Why is that so?

# Connected components of natural numbers set with subspace topology

connectednessgeneral-topology

#### Related Solutions

This was originally published as part of this note which covered all three Munkres topologies on $\mathbb{R}^\omega$ (based on the posting on topology atlas forums. For easier reference on this site I re-use the most complicated part on the box topology:

Let $X = \mathbb{R}^\omega$ be a countable product of copies of $\mathbb{R}$., and we give $X$ the box-topology.

Define a relation $\sim$ on $X$ as follows: $x \sim y$ iff the sequence $x_n - y_n$ is $0$ from a certain index onwards (or equivalently if the set of $\{n: x_n \ne y_n\}$ is finite). This is an equivalence relation: $x \sim x$ because then we have a $0$-sequence, symmetry is evident, as $-0 = 0$, and if $N_1$ is an index from which $x_n - y_n = 0$, and $N_2$ a similar one for $y_n - z_n$, then $\max\{N_1,N_2\}$ works for $x_n$ and $z_n$.

I'll show first that the classes $[x]$ are path-connected. So fix $x$. Then define for each finite subset $I$ of $\omega$ the set $X(I) = \{(y_n) : y_n = x_n \text{ for all $n$ in $\omega \setminus I$}\}$. This is homeomorphic to $\prod_{n \in I} \mathbb{R}$ by the obvious map (and by noting that on a finite(!) product the box topology and the normal product topology agree, and a finite (normal) product of copies of $\mathbb{R}$ is path-connected). Also for all such $I$, $x$ is in $X(I)$. And $[x] = \bigcup\{X(I) : I \subset N, \text{ $I$ finite}\}$, by definition. And this union is path-connected because we can always connect two points in different $X(I)$'s via their common point $x$.

Claim (to finish): if $p$ is a point not in $[x]$, so it differs with $x_n$ for infinitely many $n$, then there is a subset $Y$ such that $p \in Y$ and $Y$ clopen (i.e. closed and open) and $x \notin Y$.

This shows that no connected subset can contain both $x$ and $y$, so $[x]$ is a maximal connected subset and hence a (path-)component of $X$.

To this end: let $U_{k,n}$ be open subsets of $\mathbb{R}$ (in the $n$-th component space) such that

- $x_n \in U_{k,n}$ for all $k$ and $n$ in $\omega$,
- $p_n \not\in U_{k,n}$ unless $p_n = x_n$,
- $\operatorname{cl} U_{{k+1},n} \subset U_{k,n}$ for all $k$ and $n$ in $\omega$.

It is obvious such sets can be chosen. Define $$ Y = \{(y_n) : \text{there is a $k$ such that for infinitely many $n$ we have $y_n \not\in U_{k,n}$}\}. $$ This $Y$ does not contain $x$, as then for all $k$,$n$ we have $x \in U_{k,n}$. It does contain $p$, as $p$ differs on infinitely many places from $x$, so for every $k$ there are infinitely many $n$ such that $p_n \not\in U_{k,n}$, by (2) above.

It remains to show that $Y$ is closed and open.

$Y$ is open: let $z$ be in $Y$. Let $k$ be given as in the definition of $Y$. So there are infinitely many indices $n$ (say $n$ in $N'$) such that $z_n$ is not in $U_{k,n}$. In particular, such $z_n$ are in the complement of $U_{k+1,n}$. Then the box-topology basic open subset $\prod_{n \in \omega} O_n$ where $O_n$ is the complement $U_{k+1,n}$, if $n$ in $N'$ and $O_n$ is $\mathbb{R}$ otherwise, is contained in $Y$ and contains $z$. (It is contained in $Y$ because $k+1$ works in the definition for $Y$ for all points of this open set.)

Y is closed: let $z \not\in Y$. So for all $k$ there are only finitely many $n$ such that $z_n \not\in U_{k,n}$. Call these finite exception sets $I_k$. These sets are increasing: if $n$ is in $I_m$ then also in $I_{m+1}$. If $n$ is in neither of the $I_k$, $z_n$ must be in the intersection (over $k$) of the $U_{k,n}$. Let $f$ be some strictly increasing map from $\omega \setminus \bigcup I_k$ into $\omega$ (if this set is non-empty). Define the following open set $O = \prod_{n \in \omega} O_n$ in the box topology: if $n$ is in $I_{k+1} \setminus I_k$, for some $k \ge 0$, let $O_n$ be $U_{k,n} \setminus \operatorname{cl} U_{k+1,n}$, for $n$ in $I_0$, let $O_n$ be $\mathbb{R}$, and for $n$ in $\omega \setminus U_{k,n}$.

I claim that $O$ contains $z$ (this is quite clear), and that $O$ is disjoint from $Y$. Why is this so? Let $w$ be in $O$. Pick $k$ in $\omega$. If $w_n$ is not in $U_{k,n}$, then we want to show there can be only finitely many of these $n$. Consider the cases for $n$: $n$ can be in $I_{k+1} \setminus I_k$ only if $i+1 \le k$, so there are only finitely many $k$ that can apply. And the union of these is finite. If $n$ is in $I_0$, who cares? Only finitely many $n$ are in $I_0$. And finally, if $n$ is in $\omega \setminus \bigcup I_k$, this can only be a problem if this is set is infinite, but then $f$ grows larger than $k$ from some index $N$ onwards, and then $w_n$ is in $U_{f(n),n} \subset U_{k,n}$ (remember that the $U$'s decrease in $k$) for $n \ge N$. So in $\omega \setminus \bigcup I_k$ there can also only be finitely many $n$ with $w_n \not\in U_{k,n}$. So this shows that for every $k$, all but finitely many $n$ have $w_n \in U_{k,n}$. So $w$ is not in $Y$, and $O$ is disjoint from $Y$. So $Y$ is closed.

The topology of $\mathbb{Q}$ is the subspace topology induced from $\mathbb{R}$ and, for instance, $(x, \infty) \cap \mathbb{Q}$ is open in $\mathbb{Q}$, so everything is fine. You can write $(x, \infty) \cap A$ as $(x, \infty) \cap \mathbb{Q} \cap A$. When taking subspace topologies, you can easily check that it makes no difference if you take it in two steps $(\mathbb{R} \to \mathbb{Q}, \mathbb{Q} \to A)$ or in one step $(\mathbb{R} \to A)$. In particular, being connected is a property of the topological space $A$, regardless of what ambient space it is embedded into.

## Best Answer

Note that the topology of $\Bbb{N}$ is discrete, i.e., all its subsets are open. To see this it suffices to note that any singleton is open in the subspace topology, and then realize that any subset is a union of singletons.

Now if $X \subset \Bbb{N}$ contains $x$ as well as other points, then $X = \{ x\} \cup (X - \{x\}$) is a decomposition of $X$ into two disjoint non-empty open sets.