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I am having a difficult time understanding the statement of the inheritance principle as presented in Pugh's Real Mathematical Analysis. First, let me clarify the notation: $A_r(p)$ or$A_r p$ is the same thing and just means the $r$ - neighborhood of a point $p$ with elements exclusively from the set $A$. Here comes the statement (and proof so you have a bit more context):

Every metric subspace $N$ of $M$ inherits its topology from $M$ in the sense that each subset $V\subset N$ which is open in $N$ is actually the intersection $V=N\cap U$ for some $U\subset N$ that is open in $M$ and vice versa.

Proof: It all boils down to the fact that for each $p\in N$ we have $N_r(p) = N ∩ M_rp$. After all, $N_r(p)$ is the set of $x ∈ N$ such that $d_N (x, p) < r$ and this is exactly the same as the set of those $x ∈ M_rp$ that belong to $N$. Therefore N inherits its r-neighborhoods from $M$. Since its open sets are unions of its r-neighborhoods, $N$ also inherits its open sets from $M$.

In more detail, if $V$ is open in $N$ then it is the union of those $N_rp$ with $Nrp ⊂ V$ . Each such $N_rp$ is $N ∩ M_rp$ and the union of these $M_rp$ is $U$, an open subset of $M$. The intersection $N ∩ U$ equals $V$ . Conversely, if $U$ is any open subset of $M$ and $p ∈ V = N ∩ U$ then openness of $U$ implies there is an $M_rp ⊂ U$. Thus $N_rp = N ∩ M_rp ⊂ V$ , which shows that $V$ is open in $N$.

Here I already had trouble understanding the statement he's trying to prove. Luckily the proof sheds some more light but I am still not quite sure so here is the way I understand it:

Condition: Let $M$ be a metric space, let $N\subset M$, let $V\subset N$.

Implication: $V$ is open in $N$ $\Longleftrightarrow$ There is a $U\subset M$ which is open in $M$ such that $V = N\cap U$

  1. Did I understand the statement correctly?

My second question revolves around the application of the inheritance principle to closed sets. The book states the following:

Corrollary: Every metric subspace of $M$ inherits its closed sets from $M$.

Proof: [...]The proof of this statement consists of two words: "Take complements".

My own proof of this statement goes as follows:

Suppose $V$ is closed in $N$. Then its complement in $N$ is open, hence $V^c = N\cap U$ for some $U$ open in $M$.Hence $V = (V^c)^c = (N\cap U)^c = N^c \cup U^c$ by DeMorgan's laws. Since $\forall x\in V$ it must hold that $x\in N$ it can't be that there are any $x\in N^c$, hence it follows trivially that $V = U^c$. (Now the problematic part) Since $U$ is open in $M$ though, $U^c$ is closed in M while being fully contained in $N$, hence $V = U^c = N\cap U^c$.

  1. Here my second question already arises. At the beginning of the proof, I already said that I am considering the complement with respect to the set $N$. However, towards the end of the proof I am using $U^c$ and I don't know whether or not $U^c$ is closed/open in $M$ or closed/open in $N$. Can someone shed a light on this?

  2. Also for the backwards direction of the proof I'm pretty much lost, I don't know how to show that. I have another tentative proof ready but I am quite unhappy with this one for the same reasons:

Suppose $V = N\cap K$ for some set $K$ closed in $N$. It follows for its complement with respect to $N$ that $V^c = N^c \cup K^c$, and since it is the relative complement it follows that $V^c \subset N$. Hence by the same reasoning as in the proof before $V^c = K^c$, and since $K^c$ open it follows that $K$ is closed, so $V$ is closed.

I am aware of the flaws in this proof. Thus any detailed feedback on where my thought patterns "went wrong" and answers to any of the two questions above would be greatly appreciated!

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  1. Did I understand the statement correctly?

Yes, you did. You might not have studied topology yet, but I'll tell you what's going on here in more general language. What Pugh is showing is that the topology on $N$ induced by the subspace metric on $N$ is exactly equal to the subspace topology on $N$. The concept of "subspace topology" is a more general one, but it's the same thing as this "inherited" topology ($V$ is $N$-open iff. $V=N\cap U$ for some $M$-open $U$). In high level terms, this is saying that the subspace metric is a "good" metric because it creates the subspace topology (which topologists already view as "good").

Your proof of the analogous result is not quite right. I noticed the issue when I read:

$V=N^c\cup U^c$... it can't be that there are any $x\in N^c$, hence it follows trivially that $V=U^c$.

This suggest $U^c\subseteq N$, so that $U\supseteq N^c$. This doesn't at all need to be true, one can take $U\subsetneq N$ for examples. This was the red flag for me: we should not be able to omit the "$N$" entirely.

The issue was that the concept of complementation is relative. Does $N^c$ mean $N\setminus N$ or $M\setminus N$? $V$ is closed in $N$ if and only if its complement in $N$ is open. Let's avoid $^c$ notation and express this as $N\setminus V=N\cap U$ for some $U$ open in $M$. Then $V=N\setminus(N\cap U)$. Notice that this is not $M\setminus(N\cap U)$ which is what you were taking "$(-)^c$" to mean.

What is $N\setminus(N\cap U)$? It's just $N\setminus U$ (as interpreted in the ambient set $M$). Let's write that as $N\cap(M\setminus U)$ instead. Are we done?

For the converse, you should not take $V=N\cap K$ for some set $K$ which is closed in $N$. Then $V=K$ because $K\subseteq N$ and the proposition is trivial. We take $K$ closed in $M$. I suggest you go over your proof taking great care to avoid the use of the complement symbol $(-)^c$ (I strongly dislike this notation, if you haven't noticed already: it's too ambiguous).

Feel free to ask for further clarifications. As a general remark, when dealing with subspace topologies it's very easy to be careless with the relative notions of "open in", "closed in", and many exercises in topology textbooks are dedicated to showing that certain relative notions do and don't carry over to the subspace.

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