Nowhere dense subsets of a dense subspace are nowhere dense in the whole space and vice versa

Question

It seems, with the following lemma, the proposition at the bottom easily follows.

If $Y\subset X$ dense. Then, for every nonempty $A\subset Y$, $\text{Int}_Y \left(\text{cl}_Y A\right)=Y\cap \text{Int}_X \left(\text{cl}_X A\right)$.

Proof. Since $Y$ is dense in $X$, we have $$\text{cl}_X \left(\text{cl}_Y A\right)=\text{cl}_X A; \quad \text{cl}_Y A=Y\cap \text{cl}_XA;\quad Y-\text{cl}_Y A=Y\cap \left(X-\text{cl}_XA\right).$$

If $\text{cl}_Y A=Y$, using the first equality, the conclusion can be easily verified. Suppose $\text{cl}_Y A\subsetneq Y$. Then, the third equality implies $Y-\text{cl}_YA$ is dense in $\text{cl}_X\left(X-\text{cl}_XA\right)$. Combining this with the second equality, then $\text{cl}_Y \left(Y-\text{cl}_YA\right)=Y\cap \text{cl}_X\left(Y-\text{cl}_YA\right)=Y\cap \text{cl}_X\left(X-\text{cl}_XA\right),$ and so \begin{align*} \partial_Y \left(\text{cl}_YA\right)&=\text{cl}_YA\cap \text{cl}_Y \left(Y-\text{cl}_YA\right)\\ &=\text{cl}_YA\cap Y\cap\text{cl}_X \left(X-\text{cl}_XA\right)\\ &=Y\cap \text{cl}_XA\cap\text{cl}_X \left(X-\text{cl}_XA\right)\\ &=Y\cap\partial_X \left( \text{cl}_XA\right). \end{align*} Therefore, $$\begin{align*} \text{Int}_Y\left(\text{cl}_Y A\right) &=\text{cl}_Y A- \partial_Y \left(\text{cl}_YA\right)\\ &=Y\cap \text{cl}_XA- Y\cap\partial_X \left( \text{cl}_XA\right)\\ &=Y\cap \left(\text{cl}_XA-\partial_X \left(\text{cl}_XA\right)\right)\\ &=Y\cap\text{Int}_X \left(\text{cl}_X A\right). \end{align*}$$ $\square$

Proposition. If $Y\subset X$ dense. A subset $A\subset Y\subset X$ is nowhere dense in $X$ iff it is nowhere dense in $Y$, where $Y$ is given its subspace topology.

Proof. If $A$ is nowhere dense in $Y$, by the lemma, $Y\cap \text{Int}_X \left(\text{cl}_X A\right)=\text{Int}_Y \left(\text{cl}_Y A\right)=\emptyset$. But since $Y$ is dense in $X$, $Y\cap \text{Int}_X \left(\text{cl}_X A\right)=\emptyset$ iff $\text{Int}_X \left(\text{cl}_X A\right)=\emptyset$. Hence $A$ is also nowhere dense in $X$. The other direction is even easier, given the lemma. $\square$

Are there any mistakes? Thanks in advance. (Motivated when answering this question.)

Update:

After searching through the internet for a few days, I found the following proposition in a book called Baire Spaces, by Haworth and McCoy (1977). But it only says "a similar proof" can be given to when $Y$ is dense in $X$, without actually providing it.

Answer 1

All correct, as far as I can see.

I find the following easier to understand, but that's just personal opinion.

For any $Y$, dense or not, $(cl_X A) \cap Y = cl_Y A$. Hence $int_X(cl_X A) \cap Y \subseteq cl_Y A$ and so $int_X (cl_X A) \cap Y \subseteq int_Y(cl_Y A)$.

The other direction does use denseness. Let $U$ be any open set of $X$ such that $U \cap Y = int_Y(cl_Y A)$. As $cl_Y A \subseteq cl_X A$ we have $(X \setminus cl_X A) \cap (U \cap Y) = \emptyset$. By denseness $(X \setminus cl_X A) \cap U = \emptyset$, and so $U \subseteq cl_X A$ and $U \subseteq int_X(cl_X A)$. Therefore $int_Y(cl_Y A) \subseteq int_X(cl_X A) \cap Y$.

Answer 2

[This is posted as Community Wiki, because it is an alternative proof, not an assessment of the OP's proof.]

A subset $A$ of a topological space $X$ is nowhere dense in $X$ if and only if for every nonempty open subset $U$ of $X$ there exists a nonempty open subset $V$ of $X$ that is contained in $U$ and disjoint from $A.$

Let $Y$ be a subset of a space $X,$ with the subspace topology, and let $A$ be a subset of $Y.$ By the criterion just given, $A$ is nowhere dense in $Y$ if and only if for every open subset $U$ of $X$ such that $U \cap Y \ne \emptyset$ there exists an open subset $V$ of $X$ such that $V \cap Y \ne \emptyset$ and $V \cap Y \subseteq U \cap Y$ and $(V \cap Y) \cap A = \emptyset.$ Because $A \subseteq Y,$ the last of these three conditions reduces to $V \cap A = \emptyset.$

Suppose first that $A$ is nowhere dense in $Y$ (making no hypothesis on $Y$).

Let $U$ be any nonempty open subset of $X.$ If $U$ is disjoint from $Y,$ then $U$ is also disjoint from $A.$ On the other hand, if $U$ is not disjoint from $Y,$ then by the criterion of the second paragraph, there exists an open subset $W$ of $X$ such that $W \cap Y \ne \emptyset$ and $W \cap Y \subseteq U \cap Y$ and $W \cap A = \emptyset.$ Let $V = U \cap W.$ Then $V$ is an open subset of $X$ that is contained in $U$ and disjoint from $A,$ and we have $$ \emptyset \ne W \cap Y = (W \cap Y) \cap (U \cap Y) = (W \cap U) \cap Y = V \cap Y, $$ so $V$ is nonempty. Thus in all cases, regardless of whether $U$ is disjoint from $Y,$ there exists a nonempty open subset $V$ of $X$ that is contained in $U$ (equal to $U$ in the first case) and disjoint from $A.$ Therefore $A$ is nowhere dense in $X.$

For the converse implication, we need the hypothesis that $Y$ is dense in $X.$ That is, every nonempty open subset of $X$ has a nonempty intersection with $Y.$

Suppose that $A$ is nowhere dense in $X.$ In the subspace topology on $Y,$ every nonempty open subset is of the form $U \cap Y,$ where $U$ is an open subset of $X,$ necessarily nonempty. Because $U$ is nonempty, and $A$ is nowhere dense in $X,$ there exists a nonempty open subset $V$ of $X$ that is contained in $U$ and disjoint from $A.$ Now $V \cap Y$ is a nonempty open subset of $Y$ that is contained in $U \cap Y$ and disjoint from $A.$ Therefore $A$ is nowhere dense in $Y.$