Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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 $f:[0,1]\to X$ 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 $p$ to $0$ and the origin to $1$. Let $H=f^{-1}\big[\{p\}\big]$; clearly $H$ is closed in $[0,1]$. Let $$V=\left\{\langle x,y\rangle\in C:y>\frac12\right\}\;;$$ $V$ is an open nbhd of $p$. Suppose that $x\in H$, so that $f(x)=p$; then by continuity of $f$ there is an open interval $U$ containing $x$ such that $f[U]\subseteq V$. Being an interval, $U$ is connected, so $f[U]$ is a connected subset of $C$ containing $p$. Let $q=\langle x,y\rangle\in f[U]\setminus\{p\}$; then $x=\frac1n$ for some $n\in\Bbb Z^+$. Let $$W=\left\{\langle x,z\rangle\in f[U]:x<\frac1n\right\}\;,$$ 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 $f$.

share|cite|improve this answer
Is there any information from the last set $\{(0,1)\}$ ? – Theorem Jan 22 '13 at 20:19
Well I did find the argument on wiki, but it was not so clear in proving it is open. I got even more confused in reading that. – Scorpio19891119 Jan 22 '13 at 20:21
@Scorpio19891119: I’ll a bit more detail, but it’ll take a few minutes. – Brian M. Scott Jan 22 '13 at 20:25
@BrianM.Scott Thanks! My teacher told us to think about this from the sequence, so I wonder whether you can give me some details from the sequence viewpoint. – Scorpio19891119 Jan 22 '13 at 20:30
@Scorpio19891119 : i don't see any information from the last set ? because it is within the first set or am i not understanding something . – Theorem Jan 22 '13 at 20:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.