Exercise 17.16

Solution for $(a)$.

For $T_3$, $A$ closed $⟺A$ finite or all of $ℝ$. Since no finite set contains all of $K$, we see that $ℝ$ is the only closed set containing $K$, and so $\overline{K}=ℝ$.

For $T_5$, we claim $\overline{K}=[0,\infty )$. For $x\in [0,\infty )$, the basis elements that contain $x$ are of the form $(-\infty ,a)$ for $a>x$. Since $(-\infty ,a)\cap K\ne \varnothing$ by the Archimedean property, that is, $\forall <!--\; FUNCTION\; APPLICATION\; -->𝜖>0\exists <!--\; FUNCTION\; APPLICATION\; -->N\in \mathbb{N}$ such that $1∕N<𝜖$, $\overline{K}=[0,\infty )$ by Theorem $17.5$.

For $T_1$, $K^{\prime }=\{0\}$ by Example 17.8, and so $\overline{K}=K\cup \{0\}$.

For $T_2$, $K$ is closed since $ℝ\setminus K=(-\infty ,\infty )\setminus K$ is a basis element, and so $\overline{K}=K$.

For $T_4$, $\overline{K}=K$ since $T_4$ is finer than $T_2$, and so $ℝ\setminus K$ is still open. □

Solution for $(b)$. $T_3$ is $T_1$ since any finite point set is closed by definition of the finite complement topology. It is not Hausdorff, for if we choose $U\ni x,V\ni y$ both open, ${(U\cap V)}^c=U^c\cup V^c$ is finite, where the equality follows from De Morgan’s Laws, and so $U\cap V$ is infinite.

$T_5$ is not Hausdorff and not even $T_1$, for $ℝ\setminus \{x_0\}$ is not a union of basis elements, and so $\{x_0\}$ is not closed.

$T_1$ is Hausdorff, for if we have $x,y\in ℝ$ and $0<𝜖<|x-y|∕2$, then $(x-𝜖,x+𝜖)\cap (y-𝜖,y+𝜖)=\varnothing$. Since Hausdorff $⟹T_1$, we see that $T_1$ is also $T_1$.

Since $T_2,T_4$ are both finer than $T_1$, we see that the open sets constructed above are still open and separate $x,y$, and so $T_2,T_4$ are still both Hausdorff and thus $T_1$. □

Proof. First, suppose that $T$ satisfies the $T_1$ axiom and consider any finite subset $A$ of $X$. Then $A$ is closed in $T$ by the $T_1$ axiom so that by definition $X-A$ is open in $T$ and hence $X-A\in T$. Then $X-A\in T^{\prime }$ as well since $T\subset T^{\prime }$ so that $X-A$ is open in $T^{\prime }$. Hence $A$ is closed in $T^{\prime }$ by definition. Since $A$ was an arbitrary finite set, this shows that $T^{\prime }$ also satisfies the $T_1$ axiom.

Now suppose that $T$ is Hausdorff, and consider any two distinct points $x$ and $y$ in $X$. Then there are neighborhoods $U$ of $x$ and $V$ of $y$, both in $T$, that do not intersect since $T$ is Hausdorff. Then clearly $U,V\in T^{\prime }$ as well since $T\subset T^{\prime }$. Hence $U$ and $V$ are neighborhoods of $x$ and $y$, respectively, in $T^{\prime }$ that do not intersect. This shows that $T^{\prime }$ is Hausdorff as desired since $x$ and $y$ were arbitrary points of $X$. □

Proof. $\text{(⊂)}$ Consider any real number $x$ and suppose that $x\notin K\cup \left\{0\right\}$, hence $x\notin K$ and $x\ne 0$. Since $x\ne 0$, we have

Case: $x<0$. Then clearly the open set $(x-1,0)$ contains $x$ but does not intersect $K$ since $0<y$ for every $y\in K$, but $y<0$ for every $y\in (x-1,0)$.

Case: $x>0$. If $1<x$, then $(1,x+1)$ contains $x$ but does not intersect $K$ since $y\le 1$ for every $y\in K$, but $1<y$ for every $y\in (1,x+1)$. If $1\ge x$ it follows from the fact that $x\notin K$ that there is a positive integer $n$ where $n<1∕x<n+1$, and hence $1∕(n+1)<x<1∕n$. Then clearly the open set $(1∕(n+1),1∕n)$ contains $x$, but we also have that it does not intersect $K$. If it did, then there would be an integer $m$ where $1∕(n+1)<1∕m<1∕n$ so that $n<m<n+1$, which we know is not possible since $n+1$ is the immediate successor of $n$ in ${\mathbb{Z}}_{\text{+}}$.

Thus in all cases, there is a neighborhood of $x$ that does not intersect $K$. This of course shows that $x\notin \overline{K}$ by Theorem 17.5 part (a). We have therefore shown that $x\notin K\cup \left\{0\right\}$ implies that $x\notin \overline{K}$. By contrapositive, this shows that $\overline{K}\subset K\cup \left\{0\right\}$.

$\text{(⊃)}$ Now consider any neighborhood $U$ of $0$ so that there is a basis element $(a,b)$ containing $0$ that is a subset of $U$. Then $a<0<b$. Clearly there is an $n\in {\mathbb{Z}}_{\text{+}}$ large enough where $a<0<1∕n<b$ and hence $1∕n\in (a,b)\subset U$. Since also $1∕n\in K$, we have that $U$ intersects $K$. Since $U$ was an arbitrary neighborhood, this shows that $0$ is in $\overline{K}$ by Theorem 17.5 part (a). Since also clearly any $x\in K$ is also in $\overline{K}$, it follows that $\overline{K}\supset K\cup \left\{0\right\}$. □

Proof. First, clearly $K\subset \overline{K}$ basically by definition. Now consider any $x\notin K$. Then clearly the set $B=(x-1,x+1)-K$ is a basis element of $T_2$. Also, it clearly contains $x$ since $x\notin K$ and also does not intersect $K$ since $y\in B$ means that $y\notin K$. This shows that $x$ is not in $\overline{K}$ by Theorem 17.5 part (b). Since $x$ was arbitrary this shows that $x\notin K$ implies that $x\notin \overline{K}$. Thus $\overline{K}\subset K$ by contrapositive. This suffices to show that $\overline{K}=K$ as desired. □

Proof. Consider any real $x$ and any neighborhood $U$ of $x$. Then $U$ is open in $T_3$ so that $ℝ-U$ must be finite, noting that $ℝ-U$ cannot be all of $ℝ$ since $U$ would then have to be empty since $U\subset ℝ$, whereas we know that $x\in U$. It then follows that there are a finite number of real numbers not in $U$. However, clearly, $K$ is an infinite set so there must be an element of $K$ that is in $U$. This shows that $K$ intersects $U$ so that $x$ is in $\overline{K}$ by Theorem 17.5 part (a) since $U$ was an arbitrary neighborhood. Hence $ℝ\subset \overline{K}$ since $x$ was arbitrary. Clearly also $\overline{K}\subset ℝ$ so that $\overline{K}=ℝ$. □

Proof. Clearly $K\subset \overline{K}$. So consider any real $x$ where $x\notin K$.

Case: $x\le 0$. Then the set $B=(x-1,x]$ is clearly a basis element of $T_4$ that contains $x$. For any $y\in K$ we have that $x\le 0<y$ so that $y\notin B$. Hence $B$ does not intersect $K$.

Case: $x>0$. If $1\le x$ then it has to be that $1<x$ since $1=1∕1\in K$ but $x\notin K$, and hence $x\ne 1$. Then $B=(1,x]$ is clearly a basis element of $T_4$ and contains $x$. This also clearly does not intersect $K$ since $y\le 1$ for any $y\in K$ so that $y\notin B$. On the other hand, if $1>x$ then there is an integer $n$ where $n<1∕x<n+1$ so that $1∕(n+1)<x<1∕n$ since $x\notin K$. It then follows that the set $B=(1∕(n+1),x]$ is a basis element of $T_4$ that contains $x$ and does not intersect $K$.

Hence in all cases, there is a basis element $B$ containing $x$ that does not intersect $K$. This shows that $x\notin \overline{K}$ by Theorem 17.5 part (b). Hence we have shown that $x\notin K$ implies that $x\notin \overline{K}$, which shows by contrapositive that $\overline{K}\subset K$. Therefore $\overline{K}=K$ as desired. □

Proof. First, let $A=\left\{x\in ℝ\mid 0\le x\right\}$ and consider any $x\in A$ and any basis element $B=(-\infty ,a)$ containing $x$. Hence clearly $0\le x$ since $x\in A$ and $x<a$ since $x\in B$. Thus $0\le x<a$ so that there is an integer $n$ large enough that $0<1∕n<a$. Then $1∕n\in B$ and also clearly $1∕n\in K$. Thus $B$ intersects $K$. Since $B$ was any neighborhood of $x$ it follows from Theorem 17.5 part (b) that $x\in \overline{K}$. Hence $A\subset \overline{K}$ since $x$ was arbitrary.

Now suppose that $x\notin A$ so that $x<0$. Then the set $B=(-\infty ,0)$ is clearly a basis element of $T_5$ that contains $x$. Since $0<y$ for any $y\in K$, it follows that $y\notin B$, and thus $B$ cannot intersect $K$. Hence by Theorem 17.5 part (b) we have that $x\notin \overline{K}$. This shows that $\overline{K}\subset A$ by contrapositive, which completes the proof that $\overline{K}=A$. □

Proof. First consider any two distinct points $x,y\in ℝ$. Without loss of generality, we can assume that $x<y$. Let $z=(x+y)∕2$ so that clearly $x<z<y$. Then obviously the open intervals $(x-1,z)$ and $(z,y+1)$ are disjoint open sets in $T_1$ that contain $x$ and $y$, respectively. This shows that $T_1$ is a Hausdorff space and therefore also satisfies the $T_1$ axiom by Theorem 17.8.

It then follows that $T_2$ and $T_4$ are also both Hausdorff and satisfy the $T_1$ axiom. This follows from Lemma 1 since it was shown in Exercise 13.7 that $T_1\subset T_2\subset T_4$. □

Proof. So first consider any finite subset $A$ of $ℝ$. Let $U=X-A$ so that clearly $A=X-(X-A)=X-U$. Then, since $X-U=A$ is finite, it follows that $U$ is open in $T_3$ by the definition of the finite complement topology. Hence by definition $A$ is closed in $T_3$ since $X-A=U$ is open. This shows that $T_3$ satisfies the $T_1$ axiom since $A$ was an arbitrary finite subset of $ℝ$.

To show that $T_3$ is not Hausdorff, consider any open set $U$ containing $0$ and any open set $V$ containing $1$. It then has to be that $ℝ-U$ is finite since it cannot be that $ℝ-U=ℝ$ itself since then $U$ would have to be empty (which we know is not the case since $0\in U$) since $U\subset ℝ$. Likewise $ℝ-V$ is also finite. Thus there are a finite number of real numbers that are not in $U$ and a finite number that is not in $V$. From this, it clearly follows that there are a finite number of real numbers $x$ where $x\notin U$ or $x\notin V$. Since we have

it has to be that there are a finite number of reals numbers that are not in $U\cap V$. But since $ℝ$ is infinite, this means that there are an infinite number of real numbers that are in $U\cap V$. Hence $U\cap V\ne \varnothing$, i.e. they intersect. Since $U$ and $V$ were arbitrary neighborhoods, this shows that $T_3$ is not Hausdorff by the negation of the definition. □

Proof. First, consider the distinct real numbers $0$ and $1$. Consider then any open set $V$ containing $1$ so that there is a basis element $B=(-\infty ,a)$ that contains $1$ and is a subset of $U$. Clearly we have that $0\in B$ since $0<1<a$ and hence $0\in U$ since $B\subset U$. Since $U$ was an arbitrary neighborhood of $1$, it follows there is no neighborhood of $1$ that does not contain $0$. Hence $T_5$ does not satisfy the $T_1$ axiom by the negation of Exercise 17.15. It also then follows that $T_5$ is not a Hausdorff space by the contrapositive of Theorem 17.8. □