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.

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    $\begingroup$ 1) No non-empty subset of R at all whether open or closed or neither can be open in R^2 (assuming you mean the euclidean metric, which you never stated and do need to-- and that you interpret R to be $approx$ to $R \times \{0\} \subset R^2$ and $A\subset R$ to be $approx$ to $A \times \{0\}$). 2) No subset of R has limit point that aren't also in R. So yes every closed subset of R is closed in R^2. $\endgroup$ – fleablood Nov 24 '16 at 8:46
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    $\begingroup$ $\mathbb{R} \not\subset \mathbb{R}^2$ $\endgroup$ – Prince M Nov 24 '16 at 8:52
  • $\begingroup$ @PrinceM You're right. The wording of the question is sloppy. I'll fix it. $\endgroup$ – syeh_106 Nov 24 '16 at 9:10
  • $\begingroup$ No problem! @ syeh_106 $\endgroup$ – Prince M Nov 24 '16 at 20:50

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.

  • $\begingroup$ Thanks a lot for the confirmation. Yes, I did mean making the 1-1 correspodence between $x\in R^1$ and $(x,0) \in R^2$. $\endgroup$ – syeh_106 Nov 24 '16 at 9:01
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    $\begingroup$ @syeh_106 Just as a warning, it's safe to be accurate in math. So don't write untrue statements like $\mathbb R\subset\mathbb R^2$. If you mean to actually speak of $\mathbb \times\{0\}$, then write it as such. No need to confuse your readers more than you have to. (personally, I deduct points for sloppy writing like this, but I know of people who downritht consider the whole answer false). Also, $\mathbb R$ is written as \mathbb{R}. $\endgroup$ – 5xum Nov 24 '16 at 9:05
  • $\begingroup$ Your corrections are appreciated. I'll rephrased the question accordingly. $\endgroup$ – syeh_106 Nov 24 '16 at 9:08

Let $X$ be any space and $Y$ a subspace of $X$ and $E\subset Y.$ Then $cl_Y(E),$ the closure of $E$ in $Y,$ is $Y\cap cl_X(E).$ So $E$ is closed in $Y$ iff $E=cl_Y(E)=Y\cap cl_X(E).$

If $Y$ is closed in $X$ and $E\subset Y$ then $E$ is closed in $Y$ iff $E$ is closed in $X$: Proof:

(1). If $E$ is closed in $Y$ then $E=Y\cap cl_X(E)=(cl_X(Y))\cap cl_X(E)$, which is the intersection of two closed subsets of $X,$ so E is closed in $X.$

(2). If $E$ is closed in $X$ then $cl_Y(E)=Y\cap cl_X(E)=Y\cap E=E$, so $E$ is closed in $Y$.

To find an example of some $E\subset Y$ that is closed in $Y$ but not in $X,$ the simplest case is to let $E=Y,$ with $Y\ne cl_X(Y).$ For more diverse examples, let $p\in cl_X(Y)$ \ $Y$ and let $E$ be any subset of $Y$ such that $p\in cl_X(E).$

E.g. $X=\mathbb R,\;Y=(0,1]$, $\;E=\{1/n: n\in \mathbb N\}$ and $p=0.$


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