Below are some proofs involving totally disconnected sets. It would be great if the community can check if the first four are correct, however the last two are the hardest and ones I am stuck on.

"A topological space(X,τ)is said to be totally disconnected if every non-empty connected subset is a singleton set"

i)(X,τ) is totally disconnected if and only if for each a∈X, the connected component of a={a}.

Connected component of a is defined to be the union of all connected sets that contains a and it is the largest connected set that contains a. Forward: Assume all non empty connected subsets are singletons. Then since singletons are connected, for each a in X, the largest connected set that contains a is just {a} so the connected component of each a in X is just {a}

Backward: Assume for each a∈X, the connected component of a={a} Then there cannot be a connected set that is not a singleton, for if there is a connected set that contains more than one point, then the connected component of a point in that set would contain more than one point which is a contradiction. So all non empty connected subset of X are singletons.

ii)The set Q of all rational numbers with the usual topology is totally disconnected.

It has been previously proven that Q as a subspace of R is not connected. Furthermore if it were possible for a subset of Q with more than one point to be connected, we can always take an irrational in between any two rationals and find a separation.

iii) Indiscrete spaces with more than one point are not totally disconnected.

If an in-discrete space has two points or more, every subset is connected because it is impossible to find open disjoint open sets U and V that can separate the set since the open open sets are the entire set X and the empty set.

iv) Every subspace of a totally disconnected space is totally disconnected

Take a subset A of X and form the subspace topology. Assume A is not totally disconnected, that is assume there is a subset C in A with the subspace topology with more than one point that is connected. Since X is totally disconnected, there must be a separation of C in X. Lets call the open sets U and V that separate C in X. U and V in the subspace topology A forms a separation of C, so C cannot be connected in A, so the only connected subsets of A are the singletons.

v) Every countable subspace of R2 is totally disconnected If a subspace or R2 is countable, then every element is open so any subset with more than one point is separable.

vi) The Sorgenfrey line is totally disconnected.

Open sets in the Sorgenfrey line are [a,b) and unions those intervals. Any interval of the form [a,b) can be separated by taking a point between a and b, so those intervals are not connected in the Sorgenfrey line. Intervals of the form (a,b) are also open in the Sorgenfrey line and those can be seperated by (a,c), [c,b).

For intervals of the form [a,b] and (a,b], I am unsure how to find a separation for those.

(iv) If f is a continuous mapping of R into Q, prove that there exists a c∈Q such that f(x)=c, for all x∈R; that is, the only continuous functions from R to Q are the constant functions

Using the topological definition of continuity, pre-image of open sets are open. Constant functions are easily seen to be continuous, but how to show that there are the only continuous functions from R to Q? I can try a proof by contradiction and suppose the mapping is not constant, but it seems difficult showing how that leads to a contradiction. Also, since Q is totally disconnected, I assume we are suppose to somehow use that in the proof. The image of a connected set is connected and preserved by continuity. However R is not connected...

  • $\begingroup$ For the last one note that every continuous function maps connected subsets to connected subsets. Also your point (v) is wrong: a countable subset of $\mathbb{R}^2$ does not have to have open points, e.g. $\mathbb{Q}^2$. $\endgroup$
    – freakish
    Commented Feb 3, 2021 at 10:34

1 Answer 1


$(i)$ quite is trivial: if $X$ is totally disconnected the components can only be singletons because they are non-empty connected subsets. And if $C \subseteq X$ is non-empty connected it is a subset of some component and thus a singleton (for the reverse).

$(ii)$ needs only the second part of your argument: if $C \subseteq \Bbb Q$ and $q_1 < q_2$ exist in $C$, we can find an irrational $s$ inbetween and then $\{(\leftarrow,s) \cap \Bbb Q, (s,\rightarrow) \cap \Bbb Q\}$ disconnects ($\Bbb Q$ and) $C$. So any subset of $\Bbb Q$ with two or more points is disconnected.

$(iii)$ Any indiscrete space is connected by definition almost, so if has two or more points $X$ itself is a counterexample to being totally disconnected.

$(iv)$ follows directly from the definition as connectedness is absolute: if $C \subseteq Y (\subseteq X)$ were connected and had more than two points, it directly would contradict total disconnectedness of $X$.

$(v)$ Needs a different argument. A connected subset of $\Bbb R$ is either a singleton or an interval. Suppose $A \subseteq \Bbb R^2$ is connected and countable. Then $\pi_x[A]$ is connected but cannot be an interval as $A$ is only countable, so it is a singleton. Likewise for $\pi_y[A]$, so $A$ is a singleton. It's not true that every (or any) singleton in $A$ is open, we also have sets like $\Bbb Q^2$.

$(vi)$ uses that a set $[a,b)$ in the Sorgenfrey line $\Bbb S$ is both open and closed (clopen): open as it's basic open, closed as $b$ is not an adherence point of it (and $[a,b]$ is also closed, as it refines the Euclidean topology). Also sets like $[a,\rightarrow)$ are clopen too. So if $C$ has two or more points as a subset of $\Bbb S$, say $x < y$, pick $z$ inbetween and $\{[z,\rightarrow), (\leftarrow,z)\}$ disconnects ($\Bbb S$ and) $C$. Quite similar in idea to $\Bbb Q$ or $\Bbb P$ (the irrationals) that also have a base of clopen sets.

In general, by the same proof essentially, any $T_0$ space with a base of clopen sets (a so-called zero-dimensional space) is totally disconnected, provided it has at least two points. (we cannot omit $T_0$ due to the indiscrete example).

$(vii)$ is simpler: if $f:\Bbb R \to \Bbb Q$ is continuous, $f[\Bbb R]$ is connected (by continuity) and non-empty in $\Bbb Q$, which is totally disconnected so the image is a singleton and $f$ is constant.


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