Let $X$ be a topological and A be a nonempty subset of X.then choose the correct statement

general-topology

Let $X$ be a topological space and A be a nonempty subset of X. Then
choose the correct statement:

$1.$ $A$ is dense in $X$, if $(X\setminus A)$ is nowhere dense in $X.$

$2.$ $(X\setminus A)$ is nowhere dense , if $A$ is dense in $X.$

$3$.$A$ is dense in $X$, if the interior of $(X\setminus A)$ is empty.

$4.$The interior of $(X \setminus A)$ is empty , if $A$ is dense in $X$.

My attempt : my answer is option $1 ,2 $ and $ 4.$

Option $1$ is True. Take $A = \mathbb{Q}.$

Option $2$ is true. Same logic in option $1.$

Option $3$ is false. Take $A =[0,1]$ , $A^0 =A \ \text{interior} =(0,1) \neq \phi.$

Option $4$ is true. Take $A= \mathbb{Q}.$

Is its correct or not ?

Any hints/solution will be appreciated.

thanks u

Best Answer

  1. If $X\setminus A$ is nowhere dense, this means that $X\setminus (X \setminus A)) = A$ contains a dense and open subset, so indeed $A$ is dense. An example cannot answer this, you need an argument.

  2. If $A$ is dense, then $X\setminus A$ need not be nowhere dense, as shown by the example $A = \mathbb{Q}$.

  3. If the interior of $X\setminus A$ is empty, then $A$ is dense: $\overline{A} = X\setminus \operatorname{int}(X\setminus A))$.

  4. If $A$ is dense in $X$, the interior of $X \setminus A$ is empty; this follows by the same formula, or note that otherwise a non-empty interior of $X\setminus A$ could not intersect the dense set $A$.

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