Trying to understand why a set is both open and closed

[Mr Davis 97]

Take the subset of $\mathbb{R}$, $X = [0,1]\cup [2,3]$. Under the usual metric, the set $X$ is open and closed, according to my text. How is this the case? In particular, how is $[0,1]$ open in $X$?

 

[Orodruin]

Whether a subset is open or not depends on the topology you have on your full set. If your full set is $X$ and you have the induced topology, then $X$ is open in $X$ by definition (it must be!)

[0,1] is open in $X$ because it is the preimage of (for example) the open set (-2, 1.4) in $\mathbb R$.

$[0,1] \cup [2,3]$ is open in $X$ because it is the preimage of $\mathbb R$ itself (thereby satisfying the topology axiom that the set itself must be open).

 

[Orodruin]

Addendum: $X$ is open in the induced topology on $X$. It is not open as a subset of $\mathbb R$.

 

[WWGD]

Take the subset of $\mathbb{R}$, $X = [0,1]\cup [2,3]$. Under the usual metric, the set $X$ is open and closed, according to my text. How is this the case? In particular, how is $[0,1]$ open in $X$?

$[0,1] =(-0.1,1.1) \cap [0,1] \cup [2,3]$. This satisfies the definition of open in subspaces.

 

[fresh_42]

Take the subset of $\mathbb{R}$, $X = [0,1]\cup [2,3]$. Under the usual metric, the set $X$ is open and closed, according to my text. How is this the case? In particular, how is $[0,1]$ open in $X$?

See https://www.physicsforums.com/threa...-of-continuous-functions.960345/#post-6091723 for a bit more detailed explanation.

 

[Mr Davis 97]

$[0,1] =(-0.1,1.1) \cap [0,1] \cup [2,3]$. This satisfies the definition of open in subspaces.

I suppose I understand now how to show that something IS open or closed, but how do you show that something is not? For example, why isn't $(1/4,3/4)$ closed in $(1/2,1]$? Do I have to show that for all closed subsets $C$ of $\mathbb{R}$, that $(1/4,3/4) \not = (1/2,1] \cap C$?

 

[fresh_42]

I suppose I understand now how to show that something IS open or closed, but how do you show that something is not? For example, why isn't $(1/4,3/4)$ closed in $(1/2,1]$? Do I have to show that for all closed subsets $C$ of $\mathbb{R}$, that $(1/4,3/4) \not = (1/2,1] \cap C$?

You do it as usual. Show that there is a sequence in $(1/4,3/4)$ which converges to, say $3/4\,.$

 

[Svein]

Take the subset of $\mathbb{R}$, $X = [0,1]\cup [2,3]$. Under the usual metric, the set $X$ is open and closed, according to my text. How is this the case? In particular, how is $[0,1]$ open in $X$?

Easiest way (I think): [2,3] is closed in $X$. Since $[0, 1] = X -[2, 3]$, [0,1] is the complement in $X$ of [2,3] and so must be open.