# True/False Statements and Proofs Regarding Real Analysis and Basic Topology

I'm working on an exercise in a course of Real Analysis in which I've been given 10 statements for me to decide whether they are true or false and to then follow them up with proofs or counterexamples.

All statements are in regard to some basic topology, regarding limit points, compactness and such. I believe I've found an answer and a proof/counterexample for almost all of them but I'd like some hints and help with a few of them aswell as hopefully getting some clarity around those I am unsure of. We're working with Rudin's Principles Of Mathematical Analysis. The metric space that we're interested in for this exercise is $\mathbb R$ which might be good to know.

1. "A set $A$ is closed if and only if it is not open". Claim: False, the empty set provides a contradiction.

2. "If a set $A$ has an isolated point, $A$ can not be open". Claim: True. Consider $p\in A$ where $p$ isolated. Suppose $A$ open. Since $p$ isolated we can find a neighborhood $N$ around $p$ s.t. $N$ contains no element $q\in A$ other than $p$ since it is not a limit point. However this $N$ can not be a subset of $A$ and thus $p\in A$ is not interior $\Rightarrow$ $A$ is not open which is a contradiction.

Is it a valid proof? I'm unsure about the claim that my $N$ can't be a subset. If it only contains $\{p\}$ is it not still a valid subset and thus interior?

1. "If $A, B \subset \mathbb R$ and $A$ is open then $A\cap \operatorname{cl}(B)\subset \operatorname{cl}(A\cap B)$ where $\operatorname{cl}(E)$ denotes the closure of $E$." Claim:

I believe this should be true but I'm not sure what to do to prove it. I've been playing around with their intersection and union but I don't feel like I'm getting anywhere and I'm not sure where to start. Any help would be greatly appreciated.

1. "If $A$ is a bounded set then $S=\operatorname{sup}A$ is a limit point of $A$". Claim: False, a counterexample would be the finite set $A=\{1, 2, 3, 4, 5\}$ where $5$ is the supremum but which is not a limit point.

2. "If there exists a sequence ${p_{n}}$ with $p_{n}\in A$ s.t. $\lim_{n\to\infty}p_{n}=p$ then $p$ is a limit point of $A$". Claim: False, a counterexample would be a sequence $p_{n} = \{1, 1, 1, \ldots\}$ in the set $A=\{0, 1\}$. It converges to $1$ but $1$ can't be a limit point since a small enough neighborhood contains only itself.

I'm unsure about this one. Is the statement even false? I can't seem to find a proof that I'm happy with and neither can I find a counterexample that I'm content with.

1. "Every finite set is closed". Claim: This is true since a singleton is closed and a finite set can be constructed as the union of these singletons. Finite union of closed is closed.

2. "The set $\bigcap_{n\geq1}(-1-\frac{1}{n},1]$ is compact". Claim: This is true since the intersection of these given sets converge to $[-1,1]$ which is both closed and bounded $\Rightarrow$ it is compact.

3. "An open set that contains every rational number must be all of $\mathbb R$". Claim: False, a counterexample is the set $(-\infty,\pi)\cup (\pi,\infty)$.

4. "If $A\subset\mathbb R$ and $F$ is an open set s.t. $F\subset A$ then $F\subset \operatorname{Int}(A)$ where $\operatorname{Int}(A)$ denotes the interior of $A$". Claim: True. If $A$ open then trivial. If we suppose $A$ not open. Since $F$ open every $p\in F$ interior, thus $\exists$ neighborhood $N$ around $p$ s.t $N\subset F$. But then $N\subset F$ and therefore all $p$ interior in $A$. Thus $F\subset \operatorname{Int}(A)$.

5. "Let $A\subset\mathbb R$ be an arbitrary set and let $K\subset\mathbb R$ be compact. Then $A\cap K$ is compact.". Claim: False. Consider $A = (\alpha, \beta)$, $B = [\alpha, \beta]$. Their intersection is the set $(\alpha, \beta)$ which is open so it is not compact.

Any criticism would be appreciated. I have little to no one that I can ask whenever questions arise with my work since the course is distance and there are few opportunities and people to ask for assistance on campus. Thanks in advance!

• "If it only contains {p} is it not still a valid subset and thus interior?" It is a valid subset. But it may or may not be an open subset. If it can be open this is false. If it can not be open this is true. If you are using the euclidean metric this can not be open. if you are using the discrete metric (Or using the metric space $\mathbb Z$) it can be open. – fleablood Oct 11 '16 at 16:54
• 5) your counter example is good. isolated points are not limit points and you can have sequences converging to isolated points (provided the sequences are "eventually constant") so ... the statement is false. – fleablood Oct 11 '16 at 17:01
• "The metric space that we're interested in for this exercise is R which might be good to know." Okay, this is a subtle but important question. Or proving these statements for R with euclidean metric specifically or for metric spaces in general? – fleablood Oct 11 '16 at 17:06
• @fleablood Should probably have written that note more explicitly. What the exercise states (in the discussion leading up to the mentioning of the 10 statements) is that we consider R with the distance given by absolute value. So I assume what my teacher is saying is that we are in R1 with the Euclidean metric. – Anton Oct 11 '16 at 17:14
• Then you are good. 2) is true. But you need to explain that $\{p\} \ne N(r,p)$ for any $r$ and thus $\{p\}$ is not open. 2) can only be false if singletons can be open. In the reals with distance defined as absolute value (which what "euclidian metric" means) singletons, of course, can not be open. So is true. – fleablood Oct 11 '16 at 17:28

Pre-script: In this post I frequent point out it depends on the topology. Since originally posting we have verified that for this exercise we are assumed to be using $\mathbb R$ with the "normal" Euclidean metric (i.e. the metric where $d(x,y) = ||x - y|| = |x -y|$). So my concerns were unnecessary. I leave them in because the are important for analysis with other metrics or topologies which will occasionally require one not make assumptions.

2) Very close. But only if you are using the euclidean metric. If you are using another topology this may not be true.

Counter example. The discrete topology where all sets are open. Hence a set with isolated points is open because all sets are open because .... we said so.

If that seems breezy, consider the discrete metric where $d(x,y) = 1 \iff x \ne y$ and $d(x,y) = 0 \iff x = 0$.

Then take an isolated point (or any point) $p$ is any set $A$ and consider $B(p, 1/2) = \{x \in \mathbb R|d(x,p) < 1/2\}$. Well $B(p,1/2) = \{x \in \mathbb R|d(x,p) = 0\} = \{p\}\subset A$. So that IS and open set that IS a subset of $A$ so it is an interior point so it IS open.

However if we are using the euclidean metric or any topology with an archemedian property it is true for almost the reasons you give:

"Since p isolated we can find a neighborhood N around p s.t. N contains no element q∈A other than p"

True, because that is the definition of isolated point.

"since it is not a limit point" Well, that's kind of backwards and almost circular. It's not a limit because it is isolated. And it's isolated because it's not a limit. But never mind that's a minor point. I'd leave that clause out.

" However this N can not be a subset of A "

Why not? That's the rub. $N$ contains no other points of $A$ than $p$ and if it contains any points other than $p$ those points can not be in $A$ and so $N$ is not a subset of $A$. But what if $N$ = $\{p\}$ a there aren't any points other than $p$. Then $N$ is a subset.

Well, for $\mathbb R$ and the euclidean metric, we have the archimedian property that for any $\delta > 0$ and any point $p$ there exists a real number $y$ so that $p < y < p + \delta$.

so let $N$ be the open neighborhood of $p$ that contains no point of $A$ other than $p$. Let $\delta = radius(N)$. Then there are points, $x\in \mathbb R$ where $0 < d(p,x) <\delta$ and thus $x \in N; x \ne p \implies x \not \in A$.

And your proof is now good.

4 and 5 are good. Isolated points are not limit points yet isolated points can be supremema and/or limits of sequences. That is, if the sequence is "eventually constant" That is: if $p_n \rightarrow p$ and $p$ is isolated is $\{p_n\} \subset A$ then eventually there will be an $\epsilon > 0$ where $N(\epsilon, p)$ contains no points of $A$ and so there is an $N$ so that $n > N$ would imply $\epsilon > d(p_n,p) = 0$ i.e. $p_n = p$.

6) is interesting. It is not true for topologies in general. We can have a topology in which the empty set and the universe are the only open sets, and thus the only closed sets.

I don't think this can be false on metric spaces, however. A set with a singleton point $\{p\}$, as you point out, must be closed as there can be no limit points to the set. (If any $x \ne p$ then $d(x,p) >0$ and $N(d(x,p),x)$ contains no point of $\{p\}$ so $x$ is not limit point.) So $\{p\}$ is vacuously closed as all zero limit points are in the set.

7) is good but you may need to prove/argue that demonstrate that $\cap (-1- 1/n, 1] = [-1,1]$. I trust that will not be difficult.

The rest are all good. I'd say you did a really good job.

The third statement is true. Here is a sketch of the proof. Let $x \in A \cap \overline{B}$. Then there exists a sequence $x_n \in B$ such that $x_n$ converges to $x$. But since $A$ is open, eventually the $x_n$ are going to be elements of $A$. So deleting finitely many terms in the beginning of the sequence $x_1, x_2, ...$ you will have a sequence in $A \cap B$ which converges to $x$, so $x \in \overline{A \cap B}$. Everything else looks good.

• I'm not quite sure if I understand your proof. How can we find a sequence X_n towards x in B? Isn't that only when we know that x is a limit point? And how can we see that this sequence ever enters A? And I'm not sure what you mean by "deleting the terms in the beginning of the sequence", is it just considering the subsequence up to a certain point n and onwards. I'm sorry but I'm really not seeing the most basic explanations surrounding this problem and I'm really struggling. – Anton Oct 12 '16 at 16:32
• Let $x \in A \cap \overline{B}$. In particular, $x \in \overline{B}$. Now, an element $x$ is a member of $\overline{B}$ if and only if there is a sequence $x_1, x_2, ... \in B$ such that $\lim\limits_{n \to \infty} x_n = x$. Is this clear to you? – D_S Oct 12 '16 at 20:53
• I think so, we've seen a theorem that states that if p is a limit point in E then we can find a sequence {p_n} in E such that lim p_n = p. It isn't presented as an iff statement in my book though. – Anton Oct 13 '16 at 7:54
• In any metric space (like $\mathbb{R}$), this is true, and not difficult to prove. Now $A$ is open, with $x \in A$, so there exists an $\epsilon > 0$ such that all points in $\mathbb{R}$ whose distance from $x$ is less than $\epsilon$, must lie in $A$. – D_S Oct 13 '16 at 17:00
• But since $\lim\limits_{n \to \infty} x_n = x$, there exists an $N$ such that $|x_n - x| < \epsilon$ for all $n \geq N$. It follows that $x_{N}, x_{N+1}, x_{N+2}, ...$ are all elements of $A$. Does this make sense? – D_S Oct 13 '16 at 17:02