Exercise 36.5

Proof of $(a)$. For any $x\in X$, for $a$ large enough either $(-a,0)\cup \{p\}\cup (0,a)$ or $(-a,0)\cup \{q\}\cup (0,a)$ contains $x$. Moreover, if we have two basis elements $B_1,B_2$, their intersection is either empty, already another basis element, or a set of the form $(-a,0)\cup (0,a)$, we have a basis for $X$, for if $x\in B_1\cap B_2$, $B_1\cap B_2$, or in the last case choosing $(-a,0)$ or $(0,a)$, would be a basis element containing $x$ that is contained in the intersection. □

Proof of $(b)$. Let $f:X\setminus \{q\}\to ℝ$ be defined such that $x\mapsto 0$ if $x=p$, and $x\mapsto x$ otherwise. Clearly $f$ is a bijection; it suffices to show it is continuous and open. A basis element not containing $p,q$ maps to a basis element not containing $0$ by definition and vice versa. A basis element of the form $(-a,0)\cup \{p\}\cup (0,a)$ maps to $(-a,a)$, and in the other direction, an open interval $(-a,b)$ for $a,b>0$ maps to $(-a,0)\cup \{p\}\cup (0,a)\cup (0,b)$ if $a<b$, and similarly if $a>b$. Thus, $f$ is a homeomorphism. Note $g:X\setminus \{p\}\to ℝ$ is also a homemorphism by the same argument. □

Proof of $(c)$. To show $X$ is $T_1$, it suffices to show $\{x\}$ is closed for all $x\in X$. $\{p\},\{q\}$ are closed since $X\setminus \{p\}=(-\infty ,0)\cup \{q\}\cup (0,\infty )$, $X\setminus \{q\}=(-\infty ,0)\cup \{p\}\cup (0,\infty )$ are open. $\{x\}$ is closed for $x\ne p,q$ since if $x<0$, $X\setminus \{x\}=(-\infty ,x)\cup (x,0)\cup \{p,q\}\cup (0,\infty )$ is open, and likewise for if $x>0$. Note $X$ is not Hausdorff since any neighborhood $U$ of $\{p\}$ intersects any neighborhood $V$ of $\{q\}$, since $U,V$ must contain basis elements $B_p,B_q$ that contain $p,q$ respectively; however, $B_p\cap B_q=((-a,0)\cup \{p\}\cup (0,a))\cap ((-b,0)\cup \{p\}\cup (0,b))=(-c,0)\cup (0,c)\ne \varnothing$ for $c=\min <!--\; FUNCTION\; APPLICATION\; -->\{a,b\}$. □

Proof of $(d)$. We claim the basis elements with rational end points form a countable basis. For any $(a,b)$ basis element containing $x$, we can find $a<r<s<b$ such that $x\in (r,s)\subseteq (a,b)$; for any $(-a,0)\cup \{p\}\cup (0,a)$ basis element containing $x$, we can find $0<r<a$ such that $x\in (-r,0)\cup \{p\}\cup (0,r)\subseteq (-a,0)\cup \{p\}\cup (0,a)$. Thus they form a basis by Lemma $13.2$.

Now for any $x\ne p,q$, we see that there is a neighborhood $U$ of $x$ not containing $p,q$, and $U$ is homeomorphic to a neighborhood in $ℝ$ by $(b)$. For $x=p,q$, any neighborhood of $x$ will contain a basis element $(-a,0)\cup \{x\}\cup (0,a)$, which is homeomorphic to $(-a,a)$ by $(b)$. □