What is the smallest closed interval that contains all limit points of (0,1)?

[Askhwhelp]

1) B = (0,1) U {2}, the derived set of B which contains all limit points in B is [0,1], right?

2) B = (0,1) intersect Q,the closure of B is [0,1] too right?

3) a subset of R whose only limit point is 1... Will {1+ 1/n: n is natural number} work?

4)
an interval on which the function f(x) = 1/(1-x^2} is uniformly continuous

R \ {1} work?

 

[mtayab1994]

Number 2- the closure of B is [0,1] because it's the smallest closed set that contains B and it's limit points which are 0 and 1.

Number 4- There's a little error. Solving for x: 1-x^{2}=0\Longleftrightarrow1=x^{2}\Longleftrightarrow x=\pm1 . So in your case you'll have to choose \mathbb{R}{-1,1}

 

[Askhwhelp]

Number 2- the closure of B is [0,1] because it's the smallest closed set that contains B and it's limit points which are 0 and 1.

Number 4- There's a little error. Solving for x: 1-x^{2}=0\Longleftrightarrow1=x^{2}\Longleftrightarrow x=\pm1 . So in your case you'll have to choose \mathbb{R}{-1,1}

Will [100, 101] work for 4?

 

[jbunniii]

Will [100, 101] work for 4?

Yes, that works fine. Indeed, any compact (closed and bounded) interval on which the function is continuous will work. So you just need to avoid the two points where the denominator is zero.

 

[jbunniii]

Number 2- the closure of B is [0,1] because it's the smallest closed set that contains B and it's limit points which are 0 and 1.

In fact, 0 and 1 are not the only limit points of $B$. Every point in $[0,1]$ is a limit point of $B$.

 

[mtayab1994]

In fact, 0 and 1 are not the only limit points of $B$. Every point in $[0,1]$ is a limit point of $B$.

Yes i know, what I meant to say is that 0 and 1 or the two missing limit points so the closed interval [0,1] is the smallest closed interval that contains all of the limit points of (0,1).

 

[jbunniii]

Yes i know, what I meant to say is that 0 and 1 or the two missing limit points so the closed interval [0,1] is the smallest closed interval that contains all of the limit points of (0,1).

But $0$ and $1$ are not the only limit points missing from $B$. All of the irrationals in $(0,1)$ are also limit points of $B$ which are not in $B$. The right answer is still $[0,1]$, but it's important to recognize that $[0,1]$ is providing a lot more than two missing points.