The subset $[a, b)$ of $\Bbb R$ is neither open nor closed.

Question

While reading Topology, J.Munkres (2014), there's a following statement:

"The subset $[a, b)$ of $\Bbb R$ is neither open nor closed."

To understand this, first I have think first,

1) Is $\Bbb R$ topology? Yes. It contains empty set and itself, while closed under infite unions of its element and finit intersection of its elements.

2) However, I think the set $[a,b)$ is in $\Bbb R$ so it's open. but it says neither one of open or closed.

This apparently looks I am missing something logically, but I am short of it.

Anyone help me to understand and good start with topology?

Answer 1

"The subset $[a,b)$ of $\mathbb R$ is neither open nor closed." is a true statement.

Note that it is not open because there is no open interval around $x=a$ which is entirely included in $[a,b).$

It is not closed because its complement $$(-\infty ,a)\cup [b, \infty )$$ is not open.

Answer 2

It is not closed since $b_n=b-{{b-a}\over n}\in [a,b), n>0$ $lim_nb_n=b$ is not an element of $[a,b)$. It is not open since the intersection of every open interval whose center is $a$ with the complementary space of $[a,b)$ is not empty.

Answer 3

When author doesn't mention about topology on $\mathbb{R}$ , that means you have to consider usual or standard topology on it .

It means under usual topology on $\mathbb{R}$ , subset $[a,b)$ is niether open nor closed.

Since :

1. every open set containing $a$ is not contained in $[a,b)$ so given subset is not open in $\mathbb{R}$

2. point $b$ is limit point/accumulation point but doesn't belong to $[a,b)$ so given subset is not closed in $\mathbb{R}$

Answer 4

Assume that intervals of the form $[a,b)$ are open in $\mathbb{R}$ with $b\in\mathbb{R}$. Then since $\mathbb{R}$ is open (it is in fact both open and closed) then $\mathbb{R}\setminus[a,b)=(-\infty,a)\cup[b,+\infty)$ must be closed. But then $(-\infty,a)\cup[b,+\infty)$ can be expressed as the union of three sets $(-\infty,a)\cup[b,c)\cup(d,+\infty)$ with $b<d<c$ which is again open since all three sets are open. Therefore a contradiction. Next assume that intervals $[a,b)$ are closed with $b\in\mathbb{R}$. By definition any sequence $(x_n)\subseteq [a,b)$ if it converges its limit point is still in the interval $[a,b)$. Take $x_n:=b-1/n$ for $n\in\mathbb{N}$. You can convince yourself that for each $n$ it holds $x_n\in[a,b)$ but $\lim_nx_n=b\neq[a,b)$. Hence $[a,b)$ is not closed either.