Let \(X\) denote the rational points of the interval \([0, 1] \times 0\) of \(\mathbb{R}^2\). Let \(T\) denote the union of all line segments joining the point \(p = 0 \times 1\) to points of \(X\).
Show that \(T\) is path connected, but is locally connected only at the point \(p\).
Find a subset of \(\mathbb{R}^2\) that is path connected but is locally connected at none of its points.
Solution
For \(q\in[0,1]\cap\QQ\) write \(s_q\) for the segment from \(p=(0,1)\) to \((q,0)\), so \(T=\bigcup_qs_q\). A space is locally connected at \(z\) if every neighborhood of \(z\) contains a connected open neighborhood of \(z\).
1. \(T\) is path connected and locally connected at \(p\).
Each \(s_q\) contains \(p\), so \(T\) is star-shaped about \(p\): every \(z\in T\) is joined to \(p\) by the segment \([z,p]\subseteq T\). For \(\varepsilon>0\) the open set \(B(p,\varepsilon)\cap T\) is also star-shaped about \(p\), because \(B(p,\varepsilon)\) is convex, so it is path connected.
2. On \(T-\{p\}\) the function \(\varphi(x,y)=x/(1-y)\) is continuous and equals \(q\) on \(s_q-\{p\}\); hence every connected subset of \(T-\{p\}\) lies in a single segment \(s_q\).
The point of \(s_q\) at height \(y<1\) is \((q(1-y),y)\), and \(1-y\ne0\) on \(T-\{p\}\). The image of a connected set under \(\varphi\) is a connected subset of \(\RR\) contained in \(\QQ\), hence a single point.
3. \(T\) is not locally connected at any \(z\ne p\).
2.But the points \((q'(1-y_0),y_0)\in s_{q'}\) with rational \(q'\ne q\) and \(q'\to q\) converge to \(z\) and lie outside \(s_q\), so no such \(V\) is open in \(T\).
4. For \(k\ge0\) let \(T_k=T+(0,k)\), the union of the segments from \(a_k=(0,k+1)\) to the points \((q,k)\) with \(q\in[0,1]\cap\QQ\), and let \(Y=\bigcup_{k\ge0}T_k\). Then \(Y\) is path connected.
Each \(T_k\) is path connected by step 1, and \(a_k=(0,k+1)\) is the endpoint with \(q=0\) of a segment of \(T_{k+1}\), so \(T_k\) and \(T_{k+1}\) meet. Any two points of \(Y\) lie in \(T_j\cup T_{j+1}\cup\cdots\cup T_k\) for some \(j\le k\), a path-connected union of consecutively meeting path-connected sets.
5. \(Y\) is not locally connected at \(a_{k-1}=(0,k)\) for \(k\ge1\).
Let \(U=Y\cap B(a_{k-1},\frac12)\), so that \(U=(U\cap T_k)\cup(U\cap T_{k-1})\) and the two pieces meet only in \(a_{k-1}\). Each \(T_j\) is closed in \(Y\), because its limit points outside \(T_j\) lie on segments from \(a_j\) to irrational points \((s,j)\), which miss \(Y\); so both pieces are closed in \(U\). Let \(\psi(x,y)=x/(k+1-y)\) on \(U\cap T_k\) and \(\psi=0\) on \(U\cap T_{k-1}\). Both give \(0\) at \(a_{k-1}\), so \(\psi\colon U\to\RR\) is continuous by the pasting lemma, and its values are rational; hence \(\psi\equiv0\) on every connected subset of \(U\) containing \(a_{k-1}\). Every neighborhood of \(a_{k-1}\) contains points \((q(1-t),k+t)\) of \(T_k\) with rational \(q>0\) and small \(t>0\), where \(\psi=q\ne0\). So no connected open neighborhood of \(a_{k-1}\) lies in \(U\).
6. \(Y\) is not locally connected at any point \(z\in T_k-\{a_k\}\) other than the points \((0,k)\) with \(k\ge1\).
Write \(z=(x_0,y_0)\), so \(k\le y_0<k+1\). The segments of \(T_{k+1}\) lie in \(\{y\ge k+1\}\), and a point \((x,y)\) of \(T_{k-1}\) has \(0\le x\le k-y\). Hence if \(k=0\), or \(y_0>k\), or \(x_0>0\), a small open disk \(D\) about \(z\) satisfies \(D\cap Y=D\cap T_k\) and \(a_k\notin D\). The set \(T_k\) is a translate of \(T\), so step 3 shows that \(D\cap T_k\) contains no connected open neighborhood of \(z\).
7. Q.E.D.
6.By steps 4, 5, and 6, \(\boxed{Y}\) is path connected and locally connected at none of its points.