Exercise 22.6

Proof of $(a)$. Recall by p. 141 that it suffices to show every element in the partition, i.e., one-point sets $\{x\}$ for $x\notin K$ and $K$ itself, are closed in ${ℝ}_K$. The former are closed since ${ℝ}_K$ is $T_1$ since it is Hausdorff by Example 31.1, and the latter is closed since it is the complement of $ℝ\setminus K$. Thus, $Y$ is $T_1$.

We now show $Y$ is not Hausdorff. We claim that $p(0),p(K)$ are not separable; note $p(0)\ne p(K)$ since they are in different equivalence classes. Suppose $Y$ is Hausdorff, and let $V_1\ni p(0),V_2\ni p(K)$ be a separation in $Y$; they have open preimages $U_1=p^{-1}(V_1)\ni 0,U_2=p^{-1}(V_2)\supseteq K$ by definition of a quotient map. There then exists $(a,b)\setminus K\ni 0$ contained in $U_1$, and choosing $n\in \mathbb{N}$ such that $1∕n<b$, there exists $(c,d)\ni 1∕n$ contained in $U_2$, where we can assume $1∕(n+1)\le c$, since if not, we can take the intersection with $(1∕(n+1),d)$. Then, $(c,1∕n)\subseteq U_1\cap U_2$, and so $p((c,1∕n))\subseteq V_1\cap V_2$, which is a contradiction, and so $Y$ is not Hausdorff. □

Proof of $(b)$. By Exercise Exercise 17.13, we see that since $Y$ is not Hausdorff by $(a)$, the diagonal ${\mathrm{\Delta }}_Y\subseteq Y\times Y$ is not closed. $(p^{-1}\times p^{-1})({\mathrm{\Delta }}_Y)={\mathrm{\Delta }}_K\cup (K\times K)$, where ${\mathrm{\Delta }}_K\subseteq {ℝ}_K\times {ℝ}_K$ is the diagonal in ${ℝ}_K$. However, ${\mathrm{\Delta }}_K$ is closed by Exercise 17.13 since ${ℝ}_K$ is Hausdorff by Example 31.1, and so ${\mathrm{\Delta }}_K\cup (K\times K)$ is closed since is is the finite union of closed sets. Thus, the inverse image of the non-closed set ${\mathrm{\Delta }}_Y$ is closed, and so $p\times p$ is not a quotient map. □

Proof. First, consider a single point $U$ in the collection $Y$. If $U=K$ then clearly $p^{-1}\left(\left\{U\right\}\right)=K$, which we know is closed in ${ℝ}_K$ as was shown in Exercise 17.16 part (b). If $U\ne K$ then $U=\left\{x\right\}$ for some $x\in ℝ-K$ so that of course $p^{-1}\left(\left\{U\right\}\right)=\left\{x\right\}$ is closed in ${ℝ}_K$ since ${ℝ}_K$ was shown to satisfy the $T_1$ axiom, again in Exercise 17.16 part (b). Hence either way $p^{-1}\left(\left\{U\right\}\right)$ is closed in ${ℝ}_K$ so that $\left\{U\right\}$ must be closed in $Y$ since $p$ is a quotient map. This suffices to show that $Y$ satisfies the $T_1$ axiom since $U$ was arbitrary.

To show that $Y$ is not Hausdorff consider any neighborhood $V_0$ in $Y$ of the point $\left\{0\right\}$ and any neighborhood $V_K$ in $Y$ of the point $K$, noting that clearly $\left\{0\right\}$ and $K$ are distinct points in $Y$. Then set $U_K=p^{-1}(V_K)={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{V\in V_K}V$ so that $K\subset U_K$ since $K\in V_K$. Similarly, set $U_0=p^{-1}(V_0)={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{V\in V_0}V$ so that $\left\{0\right\}\subset U_0$ since $\left\{0\right\}\in V_0$, and hence $0\in U_0$. Since $V_0$ is open in $Y$, it follows that $U_0=p^{-1}(V_0)$ must be open in ${ℝ}_K$ since $p$ is a quotient map. It then follows that there is a basis element $B_0$ in ${ℝ}_K$ such that $0\in B_0\subset U_0$. Hence $B_0=(a,b)$ or $B_0=(a,b)-K$ for some $a<0<b$. Now, since we have $0<b$, clearly there is an $n\in {\mathbb{Z}}_{\text{+}}$ large enough that $0<1∕n<b$, and by definition $1∕n\in K\subset U_K$. Then, since $U_K$ is open in ${ℝ}_K$, there must be a basis element $B_K$ of ${ℝ}_K$ such that $1∕n\in B_K\subset U_K$. Then it must be that $B_K=(c,d)$ for some $c<1∕n<d$ since $1∕n\in K$.

Next, set $e=\max <!--\; FUNCTION\; APPLICATION\; -->\left\{1∕(n+1),c\right\}$ and $x=(e+1∕n)∕2$. Then we then have that $1∕(n+1)\le e<x<1∕n$ so that $x\notin K$. We also have $a<0<1∕(n+1)\le e<x<1∕n<b$ so that $x\in B_0\subset U_0$ regardless of whether or not $K$ is included in $B_0$ or not. Lastly, we have that $c\le e<x<1∕n<d$ so that $x\in (c,d)=B_K\subset U_K$. Therefore $x\in U_0$ and $x\notin K$ so that $p(x)=\left\{x\right\}\in V_0$ since $U_0=p^{-1}(V_0)$. Likewise we have $x\in U_K$ and $U_K=p^{-1}(V_K)$ so that $p(x)=\left\{x\right\}\in V_K$ as well. Hence $\left\{x\right\}=p(x)\in V_0\cap V_K$ so that these neighborhoods intersect. Since $V_0$ and $V_K$ were arbitrary neighborhoods of $\left\{0\right\}$ and $K$, respectively, in $Y$, this shows that $Y$ fails to be Hausdorff. □

Proof. Define the set ${\mathrm{\Delta }}_Y=\left\{(U,U)\mid U\in Y\right\}\subset Y\times Y$ to be the diagonal of $Y\times Y$. Since it was shown in part (a) that $Y$ is not Hausdorff, it follows that ${\mathrm{\Delta }}_Y$ is not closed in $Y\times Y$ by Exercise 17.13. Now set $\mathrm{\Delta }={(p\times p)}^{-1}({\mathrm{\Delta }}_Y)\subset {ℝ}_K\times {ℝ}_K$, and we claim that $\mathrm{\Delta }={\mathrm{\Delta }}_{ℝ}\cup (K\times K)$, where of course ${\mathrm{\Delta }}_{ℝ}=\left\{(x,x)\mid x\in ℝ\right\}$ is the diagonal of ${ℝ}_K$.

To show this, first consider $x\times y\in \mathrm{\Delta }={(p\times p)}^{-1}({\mathrm{\Delta }}_Y)$ so that $(p\times p)(x\times y)=p(x)\times p(y)\in D_Y$. Hence $p(x)=p(y)\in Y$ so that either $x,y\in K$ so that $p(x)=K=p(y)$, or $x,y\notin K$ so that $p(x)=\left\{x\right\}=\left\{y\right\}=p(y)$ and $x=y$. Clearly, in the former case, we have $x\times y\in K\times K$, and in the latter case $x\times y=x\times x\in {\mathrm{\Delta }}_{ℝ}$. Thus either way we have $x\times y\in {\mathrm{\Delta }}_{ℝ}\cup (K\times K)$ so that $\mathrm{\Delta }\subset {\mathrm{\Delta }}_{ℝ}\cup (K\times K)$. Now consider any $x\times y\in {\mathrm{\Delta }}_{ℝ}\cup (K\times K)$. If $x\times y\in {\mathrm{\Delta }}_{ℝ}$ then $x=y$ so that of course $p(x)=p(y)$ since $p$ is a function. On the other hand, if $x\times y\in K\times K$ then we again have $p(x)=K=p(y)$. Therefore either way we have $p(x)=p(y)$ so that $(p\times p)(x\times y)=p(x)\times p(y)=p(x)\times p(x)\in {\mathrm{\Delta }}_Y$, and thus $x\times y\in {(p\times p)}^{-1}({\mathrm{\Delta }}_Y)=\mathrm{\Delta }$. Hence $\mathrm{\Delta }\supset {\mathrm{\Delta }}_{ℝ}\cup (K\times K)$ as well so that equality has been shown.

Now, since ${ℝ}_K$ is Hausdorff (again, as shown in Exercise 17.6), it follows from Exercise 17.13 that ${\mathrm{\Delta }}_{ℝ}$ is closed in ${ℝ}_K\times {ℝ}_K$, and therefore $\overline{{\mathrm{\Delta }}_{ℝ}}={\mathrm{\Delta }}_{ℝ}$. Since $K$ is also closed in ${ℝ}_K$ (also shown in Exercise 17.16), we have that $K\times K$ is closed in ${ℝ}_K\times {ℝ}_K$ per Exercise 17.3, and so $\overline{K\times K}=K\times K$. By Exercise 17.6 part (b), we then have that

which shows that $\mathrm{\Delta }={(p\times p)}^{-1}({\mathrm{\Delta }}_Y)$ is in fact closed in ${ℝ}_K\times {ℝ}_K$. This suffices to show that $p\times p$ is not a quotient map as desired since ${\mathrm{\Delta }}_Y$ is not closed in $Y\times Y$. □