Topology: Open and Closed Sets

Question

Topology: Why do we consider the set $(2.9,3.1)$ open whereas $[2.9,3.1)$ is not considered to be open?

My understanding of open sets is that an open set is one such that for an element $x$ in a set, we can find sort of a 'neighborhood' or distance more than $0$ (doesn't matter how small, so as long as it's more than $0$) that will still be in the set.

I understand that for $[2.9,3.1)$ if the point I select is on $2.9$ then any distance ($d$) from it would fall out of the set. But won't the same apply for the $(2.9,3.1)$ case as well? After all, $(2.9,3.1)$ spans a smaller range than $[2.9,3.1)$, so it stands to reason that if a point $2.9$ falls out of $[2.9,3.1)$, it will fall out of $(2.9, 3.1)$ as well.

Answer 1

$2.9\not\in(2.9,3.1)$, so we don't run into the same problem as for $[2.9,3.1)$. There is in fact a neighborhood $N_x$ of any $x\in (2.9,3.1)$ with $x\in N_x\subset (2.9,3.1)$. In fact you could choose $N_x=(2.9,3.1)$...

Answer 2

You look like a little bit confused on this topic.

First of all, you have to specify the topology, anyway let's assume we are using Euclidean topology on $\mathbb{R}$.

Second, the set $A:=[2.9,3.1)$ is not closed in this topology:

• It is not open: because you cannot find any open ball centered in $2.9$ contained in $A$.
• It is not closed: a set is closed if and only if his complementary is open, and it's complementary is $C:=(-\infty,2.9)\cup[3.1,+\infty)$, which is not open because you cannot find any open ball centered in $3.1$ contained in $C$.

About your second sentence: $2.9 \notin (2.9,3.1)$, so you don't have to check it.

Answer 3

You should look at the definition of topological space - in which is defined what a open set is - and apply definition to your case (the other answers are the application of the definitions).