I wonder whether someone can do me a favor in proving the following union of the sets $C$ is connected but not path connected:
$[0, 1] \times \{0\}$
$\{1/n\}\times [0, 1]$
$\{(0, 1)\}$
For now, I have already proved that this union of sets is connected, but for path connected, I don't know how to prove it.
Hope for any help, thanks.
HINT: Let $C$ be the whole space. Let $p=\langle 0,1\rangle$; that point is clearly the ‘bad’ one. Suppose that there is a path $f:[0,1]\to C$ from $p$ to some other point of $C$. Clearly $f^{-1}\big[\{p\}\big]$ is closed in $C$; to get a contradiction (with the known connectedness of $C$), show that it is also open. If you get stuck, you’ll find an argument in the Wikipedia article on the comb space; your space is the deleted comb space.
Added: Assume that $f$ is a path in $C$ from $p$ to the origin; we may further assume that $f$ takes $0$ to $p$ and $1$ to the origin. Let $H=f^{-1}\big[\{p\}\big]$; clearly $H$ is closed in $[0,1]$, so it has a maximum element $a\in[0,1)$. Replacing $f$ by the map
$$[0,1]\to C:x\mapsto f\big(a+x(1-a)\big)$$
if $a>0$, we may assume that $H=\{0\}$.
Let
$$V=\left\{\langle x,y\rangle\in C:y>\frac12\right\}\,;$$
$V$ is an open nbhd of $p$, so by continuity of $f$ there is a $b\in(0,1]$ such that $f\big[[0,b)\big]\subseteq V$. Let $U=[0,b)$; $U$ is connected, so $f[U]$ is a connected subset of $V$ containing $p$.
Fix $x_0\in(0,b)$; then $f(x_0)\in V\setminus\{p\}$, so there are an $n\in\Bbb Z^+$ and a $y_0\in\left(\frac12,1\right]$ such that $f(x_0)=\left\langle\frac1n,y_0\right\rangle$. Let
$$W=\left\{\langle x,y\rangle\in f[U]:x<\frac1n\right\}\,,$$
and show that $W$ and $f[U]\setminus W$ form a separation of $f[U]$, contradicting the connectedness of $f[U]$.
Added2: To do it with sequences, let $f$ be as above. For $x\in[0,1]$ let $f(x)=\langle f_1(x),f_2(x)\rangle$, so that the functions $f_1$ and $f_2$ give the coordinates of $f(x)$ for each $x\in[0,1]$. For each $n\in\Bbb Z^+$ choose an $x_n\in[0,1]$ such that $0<f_1(x_n)<\frac1n$ and $f_2(x_n)>1-\frac1n$; this is possible because the range of $f$ contains points arbitrarily close to $p$ but not equal to $p$. (If it didn’t, it wouldn’t be connected.) The sequence $\langle x_n:n\in\Bbb Z^+\rangle$ has a convergent subsequence $\langle x_{n_k}:k\in\Bbb Z^+\rangle$; let $x$ be the limit of this subsequence. Continuity of $f$ implies that $f(x)=p$. On the other hand, for any $k\in\Bbb Z^+$ we have $f_1(x_{n_k})=\frac1m$ for some $m>n_k$. Choose $\ell\in\Bbb Z^+$ so that $n_\ell>m$; then $f(x_{n_k})$ and $f(x_{n_\ell})$ are on different ‘spikes’ of $C$. The continuous function $f$ must map the closed interval with endpoints $x_{n_k}$ and $x_{n_\ell}$ to a connected set in $C$, and you can show that any connected set in $C$ that contains points on two different ‘spikes’ must contain the portion of the $x$-axis between those two ‘spikes’. In other words, there is some $y_k$ between $x_{n_k}$ and $x_{n_\ell}$ such that $f(y_k)$ is on the $x$-axis in $C$. Use this idea to construct a sequence $\langle y_k:k\in\Bbb Z^+\rangle$ such that $f_2(y_k)=0$ for each $k\in\Bbb Z^+$, and $y_k$ lies between $x_{n_k}$ and some $x_{n_\ell}$ with $\ell>k$. Show that $\langle y_k:k\in\Bbb Z^+\rangle\to x$, while every $f(y_k)$ is on the $x$-axis and hence at least $1$ unit from $f(x)=p$; this clearly contradicts the continuity of $f$.
