It's well known that $(0,1)$ is open in $\mathbb{R}$ but is not open in $\mathbb{R}^2$, when we make the 1-1 correspondence between $x\in \mathbb{R}$ and $(x,0) \in \mathbb{R}^2$. (The usual euclidean metric is assumed.) I want to confirm if the following more general statements are true:
1) No (non-empty) open subsets of $\mathbb{R}$ can be open in $\mathbb{R}^2$?
2) Every closed subsets of $\mathbb{R}$ is closed in $\mathbb{R}^2$?
The answer to 2) appeared to be yes from Closed subset of closed subspace is closed in a metric space (X,d).
Now suppose $E\subset Y\subset X$, where $X$ is a metric space. If the above statements are true, to find an $E$ that is closed in $Y$ but not closed in $X$ (i.e. somewhat dual example to $(0,1)$ being open in $\mathbb{R}$ but not in $\mathbb{R}^2$), I suppose I must find a $Y$ that is not closed in $X$. Is the following a valid example?
3) $(0,1]$ is closed in $(0, 2]$ but is not closed in $[-1,2], \mathbb{R}$ or $\mathbb{R}^2$?
4) Lastly, a statement dual to Closed subset of closed subspace is closed in a metric space (X,d) is also true, right? (i.e. an open subset of an open metric subspace is open in a metric space.)
I'd appreciate a confirmation, refutation or comments.
Best Answer
Actually, $(0,1)$ is not a subset of $\mathbb R^2$, so it is not open because of that.
However, you are probably thinking of $\mathbb R\times \{0\}$, in which case statements $1$ and $2$ would be correct.
Proof for $1$:
Take any open set $A\subset \mathbb R\times \{0\}$ and $x\in A$. Clearly, $x=(x_1, 0)$ for some $x\in\mathbb R$.
Then, any neighborhood $U$ of $x$ in $\mathbb R^2$ contains some ball $B(x, \epsilon)$ for some $\epsilon$, but that means it also contains $y=(x_1, \frac\epsilon2)$ which is not in $A$, so $A$ is not open.
Proof of $2$:
Any closed set of a closed subspace is closed, so that's easy.
For $3$, your example is valid, yes. Even more simply, if $Y$ is not closed in $X$, you can take $E=Y$ and then $E$ is closed in $Y$, but not in $X$.
For $4$, yes, an open subset of an open subspace is open in the original space. This is true generally in any topology, since, if $Y\subset X$ is a subspace, then open sets in $Y$ are defined as intersections $A\cap Y$ where $A$ is open in $X$. This means all open subsets of $Y$ are intersections of $Y$ (open in $X$) with an open set in $X$, and are therefore open.
If you want a metric-space proof, take $A\subset Y$ and $x\in A$.
Then, since $x\in A$ and $A$ is open in $Y$, there exists some $\epsilon_1$ such that $B_Y(x,\epsilon_1)\subset A$. Also, there exists $\epsilon_2$ such that $B_X(x, \epsilon_2)\subset Y$ (because $Y$ is open in $X$).
Now take $\epsilon=\min\{\epsilon_1, \epsilon_2\}$ and you can see that $B_X(x, \epsilon)\subset A$, QED.