Prove $\mathbb{Q}$ is a path-connected topological space

I am trying to comprehend how path-connectedness of a topological space is proved in essence to solve another problem (to prove the plane $$\mathbb{R}^2$$ without a line is not path-connected) and have troubles with it.

There is a quite intuitive definition of path-connectedness: a topological space is path-connected if for any two points in that space there exists a continuous function from a compact $$[a,b]$$ to that space such that its $$f(a)$$ and $$f(b)$$ are equal to those points respectively.

Now, I've made this simple problem (stated in the title) for myself to understand how it's proved.

The way I think (although it's a hand-waving proof, I do not understand the flaw I've made):

Let's split a closet interval $$[0,1]$$ into semi-intervals like $$(x_1,x_2]$$ with $$x_2 > x_1$$ (and keep the first one starting from 0 to be a closed interval for the sake of symmetry, i.e. the first one is $$[0,x]$$). Obviously, I can split this interval into countable amount of such semi-intervals. Therefore, since between any two rational points $$a$$ and $$b$$ there're countable amount of rationals, I can construct a function mapping these semi-intervals to points in $$[a,b]\cap\mathbb{Q}$$ (e.g. I can identify each semi-interval with a rational point enclosed in between to get an enumeration and that match each semi-interval to each point in enumerated set of points in $$[a,b]\cap\mathbb{Q}$$ with some bijection $$f:\mathbb{N}\to\mathbb{N}$$). A function constructed as such is continuous because for any rational point in $$[a,b]\cap\mathbb{Q}$$ there exists a whole neighbourhood mapped to this single point. And since we can construct such a continuous function for any two rational points, Q is path-connected.

Thank you!

• $\mathbb{Q}$ is not path connected at all. How can you build a continuous function from $[a,b]$ to $\mathbb{Q}$ such that $f(a)=1$ and $f(b)=2$? By the mean value theorem $\sqrt{2}$ would have to be in the image.
– Mark
Nov 12 '19 at 11:59
• Sorry, I do not understand why $\sqrt{2}$ should be in the image since the function maps to $\mathbb{Q}$, not to $\mathbb{R}$. Could you explain where I made a mistake? Thank you! Nov 12 '19 at 12:02
• @Mark Should be intermediate value theorem instead of mean value theorem.
– edm
Nov 12 '19 at 12:08
• A function to $\mathbb{Q}$ is a function to $\mathbb{R}$. Actually, it can be proved that the only connected subsets of $\mathbb{R}$ are intervals.
– Mark
Nov 12 '19 at 12:09
• I think you should take a closer look at your argument. "I can construct a function mapping..." --- write it down. "... there exists a whole neighborhood mapped to this point" --- what's that neighborhood? In particular are the endpoints of the semi-intervals contained in these neighborhoods?
– Neal
Nov 12 '19 at 12:46

If $$I$$ is an interval of $$\mathbb R$$, then every continuous map from $$I$$ into $$\mathbb Q$$ is constant. In fact, not only $$\mathbb Q$$ is disconnected, as it is totally disconnected (that is, the only non-empty connected subsets of $$\mathbb Q$$ are the singletons).

• Well, I do not understand that any map from interval to $\mathbb{Q}$ is constant. Though it seems like a simple analysis problem. Going to prove it and then my flaw is obvious. Thank you so much! Nov 12 '19 at 12:21

It is easy to use the existence of irrational numbers to give a separation of $$\mathbb{Q}$$ in its own subspace topology. For instance, $$\mathbb{Q} = (\mathbb{Q} \cap (-\infty, \pi)) \cup (\mathbb{Q} \cap (\pi, +\infty))$$ exhibits $$\mathbb{Q}$$ as a disjoint union of two non-empty relative open sets. Hence $$\mathbb{Q}$$ is disconnected, so it cannot be path connected.

• Yes, I understand, to me it even seems quite obvious from the definition of connectedness. However, I had problems exactly with the notion of path-connectedness. As it's appeared, I just did not understand the simplest property of continuous functions :) Nov 13 '19 at 7:14

There is no simple curve $$\lambda : J \to \mathbb{Q}$$, where J is some interval by considering cardinality. "Simple" meaning $$\lambda$$ is injective. And by compactness, any curve connecting two points yields a simple curve connecting the points. So, we would have a bijection between $$J$$ (uncountable) and a subset of $$\mathbb{Q}$$ (countable), a contradiction.

• I don't think this is quite right, and if it is, it is only because you are in the Euclidean topology. There are plenty of topological spaces with a countable number of points that are still path connected. Nov 12 '19 at 19:06
• Randall I have the feeling you are right.
– ZxJx
Nov 12 '19 at 19:30
• If this is true for subspaces of the Euclidean line, it would be nice to see it worked out. Although then I think it just reduces to the Intermediate Value Theorem approach posted above. Nov 12 '19 at 19:31
• My argument relies on compactness. (mathoverflow.net/questions/214526/…) Every compact Hausdorff space of size less than the continuum is totally disconnected.
– ZxJx
Nov 12 '19 at 19:37
• Interesting..... Nov 12 '19 at 19:40