1
$\begingroup$

On a quiz I was asked whether the interval [1,1) contained the number 1, and I answered that it was true, but apparently the answer is false.

But why doesn't it? I thought that "[1,1)" means "all numbers from 1 inclusive and 1 exclusive", so since 1 is included at the start it doesn't matter if it's excluded at the end since we already included it...

$\endgroup$
  • $\begingroup$ Usually $[a,b) := \{x\in\mathbb{R}:\ a \leq x < b\}$. $\endgroup$ – Rigel Oct 19 '18 at 16:25
  • 1
    $\begingroup$ $[1,1)=\{x : 1 \leq x <1\}=\emptyset$ $\endgroup$ – Chinnapparaj R Oct 19 '18 at 16:26
  • 3
    $\begingroup$ I think, @ChinnapparajR, you mean \emptyset or \varnothing ($\emptyset$, $\varnothing$), and not $\phi$. $\endgroup$ – Namaste Oct 19 '18 at 16:28
  • 1
    $\begingroup$ Think of it this way: "$[a,b)$" means "The set of $c$ which are $\ge a$ and $<b$. It's that "and" which causes the problem. $\endgroup$ – Noah Schweber Oct 19 '18 at 16:28
  • $\begingroup$ @amWhy Sadly, some texts do use $\phi$ for the emptyset (I personally hate this, but it is a usage that occurs). $\endgroup$ – Noah Schweber Oct 19 '18 at 16:29
2
$\begingroup$

Hint: $[a,b) = \{ x \in \mathbb R : a \le x < b \}$. What real numbers $x$ satisfy $1 \le x < 1$ ?

$\endgroup$
1
$\begingroup$

[a,a] = {a}

The only real number $x$ where $a\leq x\leq a$ is $a$.

[a,a) = $\phi$ (no items)

There is no real number $x$ where $a\leq x<a$.

(a,a] = $\phi$ (no items)

There is no real number $x$ where $a< x\leq a$.

(a,a) = $\phi$ (no items)

There is no real number $x$ where $a< x< a$.

$\endgroup$
0
$\begingroup$

No need to think more. HintWhat do you say about 0≤x<0. Is there exist any real number? Think about it. You will get it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.