Munkres Topology Section 24 problem 7.

Question

Problem Let $X$ be the subspace $(\leftarrow,-1)\cup [0,\rightarrow)$ of $\Bbb{R}$. Show that the function $f:X\to \Bbb{R}$ defined by $f(x)=x+1$ if $x<-1$ , $f(x)=x$ if $x\geq 0$, is order preserving and surjective. Is $f$ a homeomorphism?

Attempt Given $f:X\to \Bbb{R}$ such that $f(x)=x+1$ if $x<-1$ , $f(x)=x$ if $x\geq 0$.

Take $ a,b\in X$ such that $a<-1,b <-1$. If $a<_Xb$ then $f(a)=a+1<_\Bbb{R} b+1=f(b)$

OTOH, Take $c,d\in X$ such that $c\geq 0$ and $d\geq 0$ if $c<_Xd$ then $f(c)=c<_\Bbb{R}d=f(d)$.

Thus $f$ is order preserving. It is easy to see that $f$ is surjective.

For $f$ is not a Homeomorphism. Is it enough to show that? $f([0,\rightarrow))=[0,\rightarrow)$

$[0,\rightarrow)$ is open in subspace topology but not open in $\Bbb{R}$. Any help or suggestion will be appreciated. Thanks!

Answer 1

You forgot to include the case where $a \in (\leftarrow, -1]$ and $b \in [0,\rightarrow)$ where we also have $a <_{\Bbb R} b$ and then $f(a)=a+1$ and $f(b)=b$ but because $a < -1 < 0 \le b$ we have $$f(a)=a+1 < 0 \le b = f(b)$$ as required too.

If we give $X$ the subspace topology wrt $\Bbb R$, which is the default, then $f$ is not open, as $[0,\rightarrow)$ is open in $X$ (it's $(-1, \rightarrow) \cap X$, so relatively open) but $f[[0,\rightarrow)]= [0,\rightarrow)$ is not open in $\Bbb R$ (As $0$ is not an interior point).

As a final observation: if we'd have given $X$ the order topology (w.r.t. the order it inherits from $\Bbb R$) then $f$ would have been an order isomorphism between $X$ and $\Bbb R$ and also a homeomorphism. So in this case the order topology is different from the subspace topology and it makes a difference for $f$. In the order topology $[0,\rightarrow)$ is not open because $0$ is not an interior point (no open interval, with endpoints in $X$ (!) sits inside it).

Answer 2

Let $U_1=(\leftarrow,-1)$ and $U_2=[0,\rightarrow)$. Then $U_1,U_2$ are open and nonempty, $U_1\cap U_2=\emptyset$, and $U_1 \cup U_2=X$. Thus $X$ is not connected. Conversely $\mathbb{R}$ is connected, so they are not homeomorphic.

Edit: for completeness here is a proof that $\mathbb{R}$ connected:

Given $U_1, U_2\subseteq\mathbb{R}$ with $U_1,U_2$ open and nonempty, $U_1\cap U_2=\emptyset$, and $U_1 \cup U_2=\mathbb{R}$, let $x\in U_1$ and $y\in U_2$. Without loss of generality, suppose $x<y$. Let: $$W=\{w\in U_1| w<y\}.$$ Note $W$ is not empty ($x\in W$) and $W$ is bounded above by $y$. Thus we may let $$z=\sup W.$$

Then $z\notin U_1$ as then it would not be an upper bound for $W$, but $z\notin U_2$ as then it would not be a least upper bound. This contradicts $U_1 \cup U_2=\mathbb{R}$.