Is $[x, x)$ an invalid interval for *all* real values of $x$?

Question

I'm writing some software at work and I believe it isn't a valid interval because it seems contradictory. To me, it reads as both an inclusion and exclusion of $x$ which seems like it would nonsense in all possible cases -- not even the empty interval would satisfy both criteria.

I wanted to ask here because I don't necessarily know enough to say for sure that this interval should be considered invalid. Rather than pulling such a statement out of relatively nowhere and asserting it in code for others, I'd like to hear from those who know more about it.

Answer 1

I would call it $\emptyset$, consistently with the notation $[a,b)=\{x\in\Bbb R\,:\, a\le x\land x<b\}$. This is an interval because it is true that $$\forall x\in\emptyset,\forall y\in\emptyset,\forall z\in\Bbb R,(x<z<y\rightarrow z\in\emptyset)$$

Answer 2

To me, it reads as both an inclusion and exclusion of x which seems like it would nonsense in all possible cases -- not even the empty interval would satisfy both criteria.

Oh! I see what you are asking. It took me a few readings to get your concern.

The definition of $[x, y)$ is $\{w\in \mathbb R| x \le w < y\}$. By that definition $[x,x) = \{w\in \mathbb R|x\le w < x\} = \emptyset$. There is no problem with that.

But I think you are assuming the following propositions:

Proposition 1: $x\in [x,y)$.

Proposition 2: $y\not \in [x,y)$

If so, then we can't have $[x,x)=\emptyset$ as that would imply $x \in \emptyset$. But we can't have $[x,x)$ be any set as we need $x \in [x,x)$ and $x \not \in [x,x)$ and that's impossible.

....

But..... proposition 1: is not true. It just isn't.

.....

By definiton: $[x,y)=\{w|x\le w < y\}$. This is non-empty if $y > x$ (and in this case $x\in \color{blue}{[x,y)}$ because $\color{blue}{x \le} x\color{blue}{< y}$) and $[x,y)$ is empty if $y \le x$ (and in these cases $x\not \in \color{blue}{[x,}\color{red}{y)}$ because $\color{blue}{x \le} x \color{red}{\not < y}$.... and because.... it's empty... we just said that....).

So a proper stating of Prop 1 would be:

Prop 1': If $[x,y)$ is non-empty, then $x\in[x,y)$. Furthermore $[x,y)$ is non-empty if and only if $y >x$.

And it this case $[x,x) = \emptyset$.

And $\emptyset=[x,x)$ satisfies both Prop 1' and Prop 2. Prop 1': If $[x,x)$ were non-empty the $x$ would be in it, but $[x,x)$ is empty and $x$ isn't in it. Prop 2: $x \not\in [x,x)=\emptyset$.