# How to show that a point is not an interior point?

I understand that in order to show that a point, $$x$$, is an interior point of some set $$A \subset B$$, where $$(B,d)$$ is a metric space you just need to show that you can have an open ball around $$x$$ which is contained in $$A$$.

I was asked a question today, which just showed I don't really understand what an interior point is.

Let $$X=(\mathbb{Q}\cap[0,3])$$ be a metric space. Define $$Y=\{y \in\mathbb{Q}:2\leq y \leq 3\}$$

Is $$2$$ an interior point of $$Y$$?

I believe that it's not. The only way I thought of perphaps proving this is by taking an arbitrary open ball which is contained in $$Y$$ and show that 2 cannot be inside it. This however seems like too much work. Is there perhaps a shorter way of thinking and proving this?

• What is X doing here ? Is it the underlying metric space ? Feb 21, 2013 at 22:06
• Yes, it is the underlying metric space and $Y \subset X$ Feb 21, 2013 at 22:10
• All you need to do is show you cannot have any ball around 2 which will completely lie in Y. It is almost trivial. Feb 21, 2013 at 22:14
• @nonlinearism What do the words "interior point of some set $\mathbf A$" mean? If it means you are assuming the metric space topology on $A$ then $B$ plays no role whatsoever in determining the open subsets of $A$, just the metric on $A$ determines that. Here if you replace $X$ by $\mathbb R$, the metric will still give a topology on $A$, but the interior of $A$ will be empty if you insist that the only open sets you are allowing $A$ are the open sets of $\mathbb R$ which do not contain any irrationals so $A$ has no interior???? Feb 22, 2013 at 4:00

Note that an open $\varepsilon$-radius ball $B(2,\varepsilon)$ around $2$ is of the form $(2-\varepsilon,2+\varepsilon)\cap\mathbb{Q}\cap [0,3]$. For any $\varepsilon>0$ take a rational point $q\in\mathbb{Q}\cap [0,3]$ with $2-\varepsilon<q<2$. Now since $q\in B(2,\varepsilon)$ and $q\notin Y$, then $B(2,\varepsilon)\not\subseteq Y$. Hence $2$ is not an interior point of $Y$.

• Can you please explain to me why it's sufficient to only consider the open ball centered at 2? Does it imply that you cannot have another open ball which contains 2? Feb 21, 2013 at 22:16
• @Adeeb: Since we have for the interior, that int$(Y)=\bigcup \{U\subseteq X:U\,\,\mathrm{is}\,\,\mathrm{open}\,\,\mathrm{and}\,\,U\subseteq Y\}$, then to show that a point $x$ is not an interior point of $Y$, it suffices to show that any open nhood of $x$ is not included in $Y$. Because if $x\in\,$int$(Y)$, then there would exist an open $U\subseteq X$ with $x\in U\subseteq Y$. Feb 21, 2013 at 22:21
• @julien. I don't think the question asks whether $2$ is an interior point of $Y$ in the subspace topology of $Y$. Every point is trivially the interior point of the whole space. When $X$ is the underlying space, then $2$ is definitely not an interior point of $Y$. Feb 21, 2013 at 22:27
• @Adeeb: Right, as I thought. To put this in simple terms, what you said in the first paragraph of your question about interiors was entirely correct. What I showed here was, that any open ball around $2$ will not be included in $Y$. Hence $2$ is not an interior point of $Y$. Feb 21, 2013 at 22:34
• @Adeeb: Also what you noted there in the first comment, that any open ball in $Y$ can not contain $2$ is true. If there was such an open ball $U\subseteq Y$ with $2\in U$, you could take a $2$-centered open ball $V$ with $V\subseteq U\subseteq Y$. Now use the above proof that $V\not\subseteq Y$ to get a contradiction. Hence any open ball contained in $Y$ does not contain the point $2$. Feb 21, 2013 at 22:40

If you are asking: interior point of $Y$ with respect to $X$'s topology, the answer is no.

A proof by contradiction is a good idea.

So assume there is $r>0$ such that $$\{y\in X \;;\; |y-2|<r\}=\mathbb{Q}\cap[0,3]\cap(2-r,2+r)\subseteq Y\subseteq [2,3].$$

By density of the rationals, we can find $y$ rational in $(2-r,2)\cap [0,3]$.

Such a number will be in the lhs set, but not in the rhs set.

Technically speaking, the question is unanswerable as you've quoted it, since it doesn't specify the metric on $X$. However, it seems reasonable to assume that $X$ is meant to be equipped with the usual Euclidean metric $d(a,b) = |a-b|$, in which case the answer is "No."

To show this, observe that every open ball in $X$ centered on $2$ is of the form $(2-r,\, 2+r) \cap X$ for some positive real number $r$, and thus contains the non-empty set $(2-r,\, 2) \cap X$ which is disjoint with $Y$. Thus, $Y$ contains no open ball centered on $2$ in $X$, and so $2$ is not an interior point of $Y$ in $X$.

Note that, per the definition you've quoted at the beginning of your post, we only need to examine open balls centered on $2$. However, if you prefer, it's easy to extend the argument above to any open balls in $X$ containing $2$ by noting that they're all of the form $(a,b) \cap X$ for some $a < 2 < b$, and thus contain the non-empty set $(a,2) \cap X$ which is disjoint with $Y$. Or we can just note that any open ball (or, indeed, any open set at all) containing a point $x$ in a metric space must include, as a subset, an open ball centered on $x$, and then apply the earlier argument above.

I am giving this as another answer because I stick by my original answer even though the other answers appear to disagree with me.

This message was tagged general topology and metric spaces. Let me point out that the definition people seem to be using is NOT the standard one for either of those concepts. Some examples: We have $\mathbb R$ is a metric space containing $X$, then $X$ is a metric space with metric induced by the metric of $\mathbb R$ with no interior because for any interval about a point $x\in X$ there is an irrational number in any ball about $x$. We now have a topological space $X$ whihc is not open in its own topology because we know some bigger space that happens to contain it. Moreover the topological definition of a function $f:U\rightarrow V$ where $U$ and $V$ are topological spaces has definition of continuity "$f$ is continuous iff $f^{-1}[S]$ is open for every open subset $S\subset V$". So the function $f:X\rightarrow {\mathbb R}$, $f(x)= x$, is not a continuous function of topological spaces?

• It seems that you have some big misconceptions here. The subset $X\subseteq\mathbb{R}$ is a metric space is its own right, and with empty interior or not in $\mathbb{R}$, it has the subspace topology $\tau_{X}=\{U\cap X:U\,\,\mathrm{is}\,\,\mathrm{open}\,\,\mathrm{in}\,\,\mathbb{R}\}$. This topology is nonempty since the usual topology of $\mathbb{R}$ is an open cover! And $X$ is open in its own topology because $X=\mathbb{R}\cap X\in\tau_{X}$, where $\mathbb{R}$ is open in $\mathbb{R}$. We are using precisely the standard definitions here. Feb 22, 2013 at 6:34
• If $(X,T)$ is a Topological space, then both $X$ and $\{\}$ are clopen. If $Y\subset X$, then $Y$ can be considered as a topological space $(Y,T_Y)$ inheriting the topology of $(X,T)$ where $T_Y=\{t\cap Y:t\in T\}$. In $(X,T)$, $Y$ can be open, closed, both, or neither, depending on $T$. However, in $(Y,T_Y)$, $Y$ is clopen. So one needs to know in which topological space we are working, $(X,T)$ or $(Y,T_Y)$.
– robjohn
Feb 22, 2013 at 7:02
• @BarbaraOsofsky: Your misconception appears to be that you think "interior point" is a relation between a point and a set. Actually it is a relation between three things, namely a point, a set, and a topological space that the set is a subset of. You need all three things to be able to speak about whether you're looking at an interior point. When we speak about things we often leave it unsaid what the topological space is, but that is just convenient shorthand for speaking about being an interior point in the only topological space that has been introduced. (...) Feb 23, 2013 at 2:40
• (...) (Or sometimes, when no topological space has been mentioned, the most "standard" topological space that the set in question is a subset of). -- In the present discussion, the question is whether the combination of the three things $2$ (a point), $Y$ (a set) and $X$ (a topological space) has the interior-point property -- that's why the OP defines $X$ at all. He's not "completely ignoring" $X$; he's defining it because one cannot speak about interior points of $Y$ without having $Y$ be a subset of a topological space. Feb 23, 2013 at 2:43
• @HenningMakholm I think you are teaching topology to an emerita professor math.rutgers.edu/people/?type=emeritus...
– user26770
Feb 23, 2013 at 11:05

Here $B=X$ and $A=Y$ in your definition of open set, and the topology is the induced topology on $Y$. Hence the $(\ 1,\ 3\ ) \cap Y$ is an open subset of $Y$ with the induced topology and it contains $2$, so $2$ is an interior point of $Y$.

• I doubt he is asking whether $2$ is an interior point of $Y$ in the subspace topology of $Y$. Trivially, every point is the interior point of the whole space. If $X$ is the underlying space, then $2$ is not an interior point of $Y$. Feb 21, 2013 at 22:29
• After all... I think the OP is talking about the topology of $X$, not $Y$. Otherwise, $Y$ is open so it is trivial. Feb 21, 2013 at 22:29
• @julien Some questions for you. Why did the OP tag the question as one of general topology and metric spaces since this interpretation says that the topology a metric space has depends on whatever topological spaces happen to contain it? And why does the OP talk about $\mathbb Q$ rather than $\mathbb R$? And of course $Y$ is open in $Y$, as is true for any topological space, and what does that have to do with whether or not $2$ or even $2.5$ is in the interior of $Y$? Feb 22, 2013 at 1:59
• @BarbaraOsofsky I think you should ask the OP, that would be easier. But for your last question: since why is open in this case, the interior of $Y$ is $Y$ and it contains $2$, and even $2.5$. Feb 22, 2013 at 2:06
• @BarbaraOsofsky No. Being consistent with what I have written so far, I would say that the interior of $A$ (with respect to $\mathbb{R}$'s topology) is empty. Now regarding the topology induced on $A$ by the metric on $\mathbb{R}$, it is the discrete topology, i.e. the topology where every subset of $A$ is open, to begin with the singletons. I think you should relax and stop telling people they don't even understand a first calculus course. Feb 22, 2013 at 4:07