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5

You’re missing the ones homeomorphic to $\big\{\varnothing,X,\{1\},\{2,3\}\big\}$; there are $3$ of those. Also, your (E) group lists one topology twice: (E1) and (E3) are the same. The question wants you to list one topology from each of the $9$ groups (including the group that I just added).


3

In (ii) and (iii) you should also account for intersections and unions of families that include the open set $X$. For (ii), for instance, let $\mathscr{U}$ be a finite family of open sets. If there is a $U\in\mathscr{U}$ such that $p\notin U$, then clearly $p\notin\bigcap\mathscr{U}\subseteq U$. If not, then $\mathscr{U}=\{X\}$, and $\bigcup\mathscr{U}=X\in\...


3

Your proof is correct, but perhaps you will find this pair of comments useful: There is no need for $X$ to be infinite. It suffices to be non-empty. Note that you have never used that $X$ is infinite. Try to rewrite the two last proofs without using RAA. This is an extremely useful technique, but RAA make these particular proofs unnecessarily long and "...


3

Hint: To find a counterexample for the second question, try considering topologies on a set $X$ which has two points.


2

Given $E$ of infinite measure and $\epsilon > 0$, let $V$ and $F$ be as in (a). Then $\mu(E-F) < \epsilon$. For each $n$, put $$ F_n = F\cap \left(\bigcup_{i=1}^{n}K_i\right), $$ $F_n$, as a closed subset of a compact set, is itself compact. We have $$ \lim_{n\to \infty} \mu(F_n) = \mu(F) = +\infty, $$ from which we conclude that $\mu$ is regular.


2

Let $E\in\mathfrak{M}$. If $\mu(E)<\infty$ then $E$ is inner regular by $(d)$. Suppose $\mu(E)=\infty$ and $E$ $\sigma$-finite. To show: $E$ is inner regular. My idea Since $E$ $\sigma$-finite, there exists a countable family $\{A_n\}_{n\in\mathbb{N}}\subset \mathfrak{M}$ such that $E=\bigcup_{n\in\mathbb{N}}A_n$ and $\mu(A_n)<\infty$ for all $n\in \...


2

I lied when I said the two definitions were equivalent; I was overlooking something. Say $A$ and $B$ can be "separated by a continuous function" if there exists $f$ so that $f=0$ on $A$ and $f=1$ on $B$. Then it's true that a space is normal if and only if any two disjoint closed sets can be separated by a continuous function. Saying $A$ and $B$ are ...


2

In his Principles of Mathematical Analysis, Rudin defines "neighborhoods" as what we commonly call "open balls", and denoted $B(x,r)$ (Rudin also adopts this terminology in his Real and Complex Analysis). The standard definition of a neighborhood $N$ of $x$ is a set with an open subset $V$ such that $$ x\in V \subset N. $$ A more restrictive definition would ...


2

Given Rudin's definition of "neighborhood," if $x\in N_r(p)$ but $x\not=p$, then $N_r(p)$ is not, itself, a neighborhood of $x$. Strictly speaking, a set that is a neighborhood of a point is a ball with that point as its center, so $N_r(p)$ can only be a neighborhood of $p$. What Rudin therefore needs to prove -- and where metric considerations enter in -- ...


2

\begin{align*} X_1\times\varnothing\neq&\,X_1\text{, nor }X_1\times X_2;\\ X_1\times\varnothing=&\,\varnothing. \end{align*}


1

It seems like you have two questions, one of which I addressed in a comment. The other is how to justify the statement Since $\lim_{n\to \infty} f_n(x)=f(x)$ for all $x\in M$, it follows that $\cap_{n=1}^{\infty} F_n =\emptyset$ Note that $$ \bigcap_{n=1}^\infty F_n = \left\{x \in M : |f_n(x) - f(x)| \geq \epsilon \text{ for all } n\right\} $$ ...


1

The problem with your proof is that you're assuming that $N$ is a "neighborhood" of $x$ for every $x\in N$. That is not actually the case with the (slightly non-standard) definitions you quote. For example, on the real line, $N=(1,5)=N_2(3)$ is a "neighborhood" of $3$, and $2\in N$, but $(1,5)$ is not a "neighborhood" of $2$. So you cannot, given these ...


1

If I understand the question the answer is yes. I take it we're talking about an "extended metric", by which I gather we mean a function $d:X\times X\to[0,\infty]$ that satisfies all the axioms for a metric. Assuming that's what we mean, the reason people don't talk about extended metric spaces is that an extended metric space is a metric space. If $d$ is ...


1

Since $E$ is $\sigma$-finite, we can write $E = \cup E_n,$ where $E_1 \subset E_2 \subset E_3 \cdots,$ and $\mu(E_n) <\infty$ for each $n.$ By the result already known, for each $n$ we can choose a compact $K_n \subset E_n$ such that $\mu(E_n\setminus K_n) < 1/n.$ We then have $$\mu(E) = \lim_{n\to \infty} \mu(E_n) = \lim_{n\to \infty} \left (\mu(K_n) ...



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