Let $X=\{(x,0)\in \mathbb{R}^2\,|\,x \in \mathbb{R}\} \cup \{(0,1)\}$, give $(0,1)$ the name "$a$", and let $T=\{A\subseteq X\,|\,a \in A\} \cup \{\emptyset\}$. Is $(X,T)$ path-connected?

I believe it is, and here is why:

Let $f:[0,1] \to X$ be such that $f(0)=x$ and $f(1)=a$, where $x \in X\setminus \{a\}$ (of course the constant map to $a$ is continuous). Then $f([0,1])=\{a\}\cup X_0$ where $X_0$ is closed (does not contain $a$). Then we have $f^{-1}(\{a\}\cup X_0) = f^{-1}(\{a\}) \cup f^{-1}(X_0)$. Since $\{a\}$ is open we must have $f^{-1}(\{a\})$ of the form $(a_0,1]$ (this is open in the subspace topology). Since $X_0$ is closed we must then have $f^{-1}(X_0) = [0,a_0]$. So $f^{-1}(\{a\} \cup X_0) = [0,1]$. I don't see any contradiction in this, so is it true that $X$ is path-connected?

I worked this out because intuitively the space seems not to be path-connected but is due to the topology (if I'm correct). Is there a non-discrete topology on $X$ that makes it non-path-connected?

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    $\begingroup$ Of course there's a non-discrete topology on $X$ which makes it path-disconnected: the subspace topology from $\mathbb R^2$. Although for any set with more than $2$ points there is a non-discrete topology which makes the set disconnected (hence path-disconnected). $\endgroup$ – Alex Becker Sep 21 '12 at 3:39
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    $\begingroup$ $\bf R$ is not a subset of ${\bf R}^2$. Perhaps when you wrote $\bf R$ you meant the set of points of the form $(r,0)$. $\endgroup$ – Gerry Myerson Sep 21 '12 at 3:52
  • $\begingroup$ First sentence edited according to comments. $\endgroup$ – Alex Petzke Sep 21 '12 at 12:58

I’m going to assume that ‘$\subseteq\Bbb R^2$’ in the definition of $X$ was simply an error. (If you really want $X$ to be a subset of the plane, follow Gerry Myerson’s suggestion and replace $\Bbb R$ by $\{\langle x,0\rangle:x\in\Bbb R\}$.)

Suppose that $f:[0,1]\to X$ is continuous and $f(1)=a$. Let $A=\{t\in[0,1]:f(t)=1\}$; the set $\{a\}$ is open in $X$, so $A$ must be open in $[0,1]$. Now suppose further that $f(0)=x\ne a$, and let $Z=\{t\in[0,1]:f(t)=x\}$. The set $\{x,a\}$ is open in $X$, so $f^{-1}[A]=Z\cup A$ must be open in $[0,1]$. Suppose that we let $f(t)=a$ for all $t\in(0,1]$, so that $Z=\{0\}$ and $A=(0,1]$; that definitely makes $f$ continuous at $0$ and at $1$, and it’s easy to check that for any open set $U$ in $X$, $f^{-1}[U]$ is either $\varnothing$ (if $U\varnothing$), $(0,1]$(if $x\notin U$), or $[0,1]$ (if $x\in U$). Thus, $f$ is continuous, and you’ve shown that there’s a path from $x$ to $a$. It follows easily by concatenating paths that $X$ is path-connected.

As was pointed out in the comments, if you replace $\Bbb R$ by the $x$-axis and topologize $X$ as a subspace of $\Bbb R^2$, you get a very nice non-discrete topology on $X$ in which $X$ is not path-connected.

  • $\begingroup$ So what was wrong with what I had, if anything? $\endgroup$ – Alex Petzke Sep 21 '12 at 13:19
  • $\begingroup$ @Alex: Not a thing. I should have said so, but I dashed that off in a hurry and forgot. $\endgroup$ – Brian M. Scott Sep 21 '12 at 19:38
  • $\begingroup$ Got it. Thanks for the answer. $\endgroup$ – Alex Petzke Sep 21 '12 at 21:17

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