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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!

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    $\begingroup$ $\mathbb{Q}$ 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. $\endgroup$
    – Mark
    Nov 12 '19 at 11:59
  • $\begingroup$ 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! $\endgroup$
    – mur_tm
    Nov 12 '19 at 12:02
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    $\begingroup$ @Mark Should be intermediate value theorem instead of mean value theorem. $\endgroup$
    – edm
    Nov 12 '19 at 12:08
  • $\begingroup$ 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. $\endgroup$
    – Mark
    Nov 12 '19 at 12:09
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    $\begingroup$ 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? $\endgroup$
    – Neal
    Nov 12 '19 at 12:46
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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).

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  • $\begingroup$ 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! $\endgroup$
    – mur_tm
    Nov 12 '19 at 12:21
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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.

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  • $\begingroup$ 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 :) $\endgroup$
    – mur_tm
    Nov 13 '19 at 7:14
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Randall
    Nov 12 '19 at 19:06
  • $\begingroup$ Randall I have the feeling you are right. $\endgroup$
    – ZxJx
    Nov 12 '19 at 19:30
  • $\begingroup$ 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. $\endgroup$
    – Randall
    Nov 12 '19 at 19:31
  • $\begingroup$ My argument relies on compactness. (mathoverflow.net/questions/214526/…) Every compact Hausdorff space of size less than the continuum is totally disconnected. $\endgroup$
    – ZxJx
    Nov 12 '19 at 19:37
  • $\begingroup$ Interesting..... $\endgroup$
    – Randall
    Nov 12 '19 at 19:40

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