I would like to show the following theorem !
Let $(X,\mathcal{T})$ be a topological space and $A$ a subset of the latter. Then
- A point $x$ is in $\bar{A}$ if and only if every neighborhood of $x$ intersects $A$.
- The set $A$ is dense in $X$ if and only if $A$ intersects any non empty open set of $X$.
My attempt :
For the "only if" part we suppose that $x$ is not in $\bar{A}$, it means that $x\in X\setminus\bar{A}$ and we have found a neighborhood of $x$ (namely the open set $X\setminus\bar{A}$) which does not intersect $A$ since $A\subset\bar{A}$. For the "if" part we suppose that $V$ is an open set which contains $x$ and that does not intersect $A$, we conclude by the fact that $\bar{A}\subset X\setminus V$ since the complementary set of $V$ is a closed set in $X$ and this closed set must contain $\bar{A}$ which is the smallest closed set containing $A$.
For the "if" part, we have that $A$ is dense in $X$ which means that $\bar{A} = X$. By the first part of the theorem we know that any neighborhood of all points (that is any non empty set) in $\bar{A} = X$ intersects $A$. For the "only if" part, we know that $A$ intersects any non empty set of $X$, consider $x\in X$, every neighborhood of $x$ (which is a non empty set of $X$) intersects $A$, which implies that $x\in\bar{A}$.
Insight of the attempt :
- A neighborhood of $x$ can be simply an open set containing $x$.
- A non empty set in a topological space can always be identified to a neighborhood of a point in this space. Not clear and probably false
I would like to know if my intuition/understanding of these concepts seem correct and if my proof is coorect please ?
Thank you a lot !
You seem to be using "any non empty set" to mean "any non empty open set" but otherwise looks good to me.