Exercise 2.4.6 (Half-open interval topology)

Question

Let $\mathscr{B}$ be the family of subsets of $\mathbb{R}$ of the form $[a, b)$, where
$$
-\infty<a<b<\infty 	ext {. }
$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathscr{B}$ is a base of open sets for a topology $\mathscr{T}$ of $\mathbb{R}$. The topology determined by $\mathscr{B}$ is the half-open interval topology.
\item Show that every open subset of $\mathbb{R}$ (in the metric topology) is $\mathscr{T}$-open.
\item Show that each interval $[a, b)$ is $\mathscr{T}$-closed.
\item Show that a point $t \in R$ lies in the $\mathscr{T}$-closure of a subset $S$ of $\mathbb{R}$ if and only if there is a sequence $\left\{t_{n}ight\}_{n=1}^{\infty}$ in $S$ such that $t_{n} \geq t$ and $\left|t_{n}-tight| ightarrow 0$
\item Show that a function $f$ from $(\mathbb{R}, \mathscr{T})$ to $\mathbb{R}$ is continuous if and only if $f$ is continuous from the right at each $t \in \mathbb{R}$, that is, if and only if
$$
\lim _{\varepsilon>0, \varepsilon ightarrow 0} f(t+\varepsilon)=f(t), \quad t \in \mathbb{R} .
$$
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item extit{$\mathscr{B}$ is a base of open sets for a topology $\mathscr{T}$ of $\mathbb{R}$.} \begin{proof} One implication of \href{./exercise_2-4-5}{the previous exercise} is that a family of sets is a base for a topology iff it is closed under finite intersections. Thus, let $[a_1,b_1)$ and $[a_2,b_2)$ be two half-open intervals. Brute-force case consideration results in two possible outcomes: either $[a_1,b_1) \cap [a_2,b_2)$ is another half-open interval or it is the empty set. In both cases, the intersection is contained in $\mathscr{B}$, which inductively implies the closure under finite intersections. \end{proof} \item extit{Every open subset of $\mathbb{R}$ (in the metric topology) is $\mathscr{T}$-open.} \begin{proof} Since any open subset of $\mathbb{R}$ can be written as a union of open intervals, it suffices to demonstrate that every interval of the form $(a,b)$ is open in the half-open topology. This follows from the fact that $(a,b)$ can be written as $$(a,b) = \bigcup [a-\epsilon,b)$$ i.e., as a union of open sets in the half-open interval topology. \end{proof} \item extit{Each interval $[a, b)$ is $\mathscr{T}$-closed.} \begin{proof} Consider the complement $$\mathbb{R} \setminus [a,b) = (-\infty,a) \cup [b, + \infty) = (-\infty,a) \cup \left(\bigcup_{n=1}^{\infty} [b, b+n) ight).$$ The first set is open by the previous part and the latter is open by axiom (1.2). Thus, the complement of $[a,b)$ is open as a union of open sets. \end{proof} \item extit{A point $t \in R$ lies in the $\mathscr{T}$-closure of a subset $S$ of $\mathbb{R}$ if and only if there is a sequence $\left\{t_{n} ight\}_{n=1}^{\infty}$ in $S$ such that $t_{n} \geq t$ and $\left|t_{n}-t ight| ightarrow 0$.} \begin{proof} Suppose that $t$ lies in the closure of $S$. Since $t$ is an adherent point of $S$, every one of its open neighbourhoods meets $S$. In particular, $\left[t,t + \frac{1}{n} ight) \cap S eq \emptyset$ for all $n \in \mathbb{N}$. By \href{https://en.wikipedia.org/wiki/Axiom_of_countable_choice}{the axiom of choice}, we can construct a sequence $\{t_n\}_{n=1}^{\infty}$ by picking a $t_n$ out of each $\left[t,t + \frac{1}{n} ight) \cap S$. $\{t_n\}_{n=1}^{\infty}$ then obviously converges to $t$. Conversely, if $t$ is a limit of a sequence $\left\{t_{n} ight\}_{n=1}^{\infty}$ in $S$ such that $t_{n} \geq t$, then every open neighbourhood of $t$ is ought to contain an element of $\left\{t_{n} ight\}_{n=1}^{\infty}$. In other words, $t$ is an adherent point of $S$. \end{proof} \item extit{A function $f$ from $(\mathbb{R}, \mathscr{J})$ to $\mathbb{R}$ is continuous if and only if $f$ is continuous from the right at each $t \in \mathbb{R}$, that is, if and only if $$ \lim _{\varepsilon>0, \varepsilon ightarrow 0} f(t+\varepsilon)=f(t), \quad t \in \mathbb{R} . $$} \begin{proof} Suppose that $f$ is $\mathcal{T}$-continuous. Let $t \in \mathbf{R}$ and $\varepsilon>0$, and let $U$ be the open interval $(f(t)-\varepsilon, f(t)+\varepsilon)$. Then $f^{-1}(U)$ is $T$-open, so there is $\delta>0$ such that $f^{-1}(U)$ contains $[t, t+\delta)$. Thus, $|f(s)-f(t)|<\varepsilon$ for $t0$ such that $f(s) \in U$ for $t

Exercise 2.4.6 (Half-open interval topology)

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Exercise 2.4.6 (Half-open interval topology)

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