New answers tagged

1

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


0

Note that $(x,y)\sim(x',y')$ implies $$x^2+y^2 = (x')^2 + y^2 = 1, $$ from which $x^2 =(x')^2$. Let $\pi:S^1\to S^1/\sim$ be the canonical projection map, then $$\pi(x,y) = \{(x,y),(-x,y) \}. $$ Now if $\pi(x,y)\ne \pi(x',y')$, we have $y\ne y'$, so as $S^1$ is Hausdorff as a closed subspace of $\mathbb R^2$, there exist disjoint neighborhoods $U, U', V, V'$ ...


0

Take the diagonal $\;\Delta\;$ in the quotient space: $$\Delta=\left\{\,\left((x,y),\,(x,y)\right)\in X/\sim\,\right\}$$ and suppose we have a sequence $\;\left\{\,((x_n,y_n),(x_n,y_n))\,\right\}\subset\Delta\;$ that converges in $\;X/\sim\;$. We want to prove the limit is also in $\;\Delta\;$...but this is trivial since any set of representatives of the ...


3

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).


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 $$ \bigcap_{n=1}^\infty F_n = \left\{x \in M : |f_n(x) - f(x)| \geq \epsilon \text{ for all } n\right\} $$ ...


0

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 $m(E_n\setminus K_n) < 1/n.$ We then have $$\mu(E) = \lim_{n\to \infty} \mu(E_n) = \lim_{n\to \infty} \left (\mu(K_n) +\...


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\...


0

@ajotatxe, here is my attempt to prove (ii) and (iii) without RAA. (ii) $\mathscr{T}$ is closed under finite intersections. Let: $ A = \bigcap_{i=1}^n U_i ; $ where $U_i$ are sets that contain $p$ and $n \in \mathbb{N}$. $A$ must be in $\mathscr{T}_3$ because intersections are the points that sets have in common, and every $U_i$ contains $p$. (iii)...


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 "...


1

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


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 -- ...


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 ...


0

You can only define the neighborhood of a point or a subset so the fact you want to show does not make sense. N is a neighborhood of p if there exists an open subset which contains p and which is contained in N.


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 ...


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 \...


1

In topological spaces, a cell complex is usually the same thing as a CW-complex. The name "cell complex" comes from the fact that there exists generalizations to other categories, but if you're interested in topological spaces then for all intents and purposes "cell complex" = "CW-complex". A finite cell complex is a cell complex that has a finite number of ...


1

The subbase is a union of two families: the family of left intervals, and the family of right intervals. So the proof is correct. If your interpretation would hold, you would write it as $$S = \{(-\infty,b) \cup (a, \infty): a,b \in \mathbb{N} \}$$ which is different: a family of unions, parametrised by 2 parameters, and it would be only one family. As ...


1

You are confusing $$\{(-\infty, b) \mid b \in \mathbb{N}\} \cup \{(a,\infty) \mid a \in \mathbb{N}\}$$ with $$ \{ (-\infty, b) \cup (a, \infty) \mid b \in \mathbb{N}, a \in \mathbb{N} \}$$


1

I cannot evaluate your proof, but a "normal quasi-uniformity" is definitely not the same as the topology from the uniformity being normal. It's true in general that terms like "normal", "regular" etc. are overused. There is also a notion of normal covers, normal family etc., all of which are unrelated. The linked-to article shows that every topology of a ...


1

De Morgan's laws if you want to be specific: $$M \setminus (\cup_{i \in I} A_i) = \cap_{i \in I} (M \setminus A_i) $$ $$ M \setminus (\cap_{i \in I} A_i) = \cup_{i \in I} (M \setminus A_i) $$ when $A_i, i \in I$ are subsets of $M$. The first one with $A_\lambda$ for $A_i$ for covering $A_\lambda$ is $\emptyset$ on the left hand side, and so the ...


1

I'm looking for the largest radius of an open interval centered at $x$ and contained in $S$, I want an $\epsilon$ that does not extend past my boundary points $\{4,9\}$ Since my boundary points are in $S$, I will use them to determine the largest radius. Therefore I need to meet the following two conditions: (1) $x-\epsilon\ge4$ (2) $x+\epsilon\le9$ ...


4

This appears to be an error in Kelley. Let $X$ be an infinite set with a distinguished element $x\in X$. Consider the topology $\{\varnothing\}\cup \{U\subseteq X\mid x\in X\}$ on $X$. (I learned this example from this answer, where it is called the "particular point topology"). $X$ is not compact: the open cover $\{\{y,x\}\mid y\in X\}$ has no finite ...


2

An arbitrary set $A$ does have some natural topologies, but they're not very interesting. Nevertheless, these uninteresting topologies can be the restriction, via the constant-functions embedding, of interesting topologies on $A^X$. In particular, if $X$ isn't compact, then you can use the discrete topology on $A$ and get a non-discrete topology as the ...


0

Hint: You want to show that $f(U)$ is open in $N$ for all open $U$ in $M$. Note that as $f$ is continuous, for all $x\in U$, there are coordinate charts $U_x \subset U$ so that $f(U_x)$ lie in a coordinate chart $V_x$ of $N$. And you have $$U = \bigcup_{x\in U} U_x$$


0

Take a class $[\alpha] \in \pi_1(X,x_0)$. How many lifts of $\alpha$ exist in $\pi_1(\tilde{X},\tilde{x_0})$? This might be useful https://en.wikipedia.org/wiki/Covering_space#Lifting_properties


0

For the uniform topology: It's easy to see that $S\subset T:=\{\overline{x}\in \Bbb{R}^{\omega}\ |\ \overline{x}=(x_n)_\omega\ \mathrm{and\ } \forall \varepsilon>0\exists N\in \Bbb{Z}_+ \forall n\geq N :|x_n|<\varepsilon\}$, (i.e., the set of all real sequences that converge to $0$.) If $y\notin T$ , then there is $\varepsilon>0$ such that for ...


1

What you’ve done is fine apart from a small notational glitch: $f[C]$ is by definition the set of values $f(y)$ for $y\in C$, so it’s $\{1\}$, not $1$. And if you want to talk about $f(x)\cap f(C)$, you need to view $f(x)$ as a set as well, which means that you’re really looking at $f[\{x\}]$, the image of the set $\{x\}$.


1

The proof of your claim comes down to proving whether $\pi_X\left ( \bigcup_iW_i\right ) = \bigcup_i \pi_X(W_i)$ where the $W_i$ are product open sets. Clearly if $U\times V$ is a product open set in $X\times Y$ then $\pi_X(U\times V) = U$. $(\Longrightarrow)$ Let $p \in \pi_X\left ( \bigcup_i U_i \times V_i\right )$, then there exists an element $q \in ...


2

Since $g$ and $h$ are continuous, $g^{-1}(U)\cap h^{-1}(V)$ is an open set containing $x$ and hence it contains an element $a$ of $A$. But $a\in g^{-1}(U)\cap h^{-1}(V)$ implies that $f(a)=g(a)\in U$ and $f(a)=h(a)\in V$ which contradicts the assumption that $U,V$ are disjoint


-1

Hints: (for how to construct your own counterexamples) Use the discrete metric on the correct space. Or, if you're feeling ambitious, consider the infinite dimensional vector space $\ell^\infty$. Consider a certain compact subset of $\Bbb R^2$. There exists a non-compact space which is null-homotopic.


3

Consider the subbasic open set $$S_{V,n}=\prod_{k\in\Bbb Z^+}U_k\;,$$ where $U_n=V$ is an open set in $\Bbb R$, and $U_k=\Bbb R$ for $k\ne n$. It’s not at all hard to calculate $f^{-1}[S_{V,n}]$: $f(t)\in S_{V,n}$ if and only if $nt\in V$. If we let $g:\Bbb R\to\Bbb R$ be the function $g(t)=nt$, we can express this in terms of $g$: $t\in f^{-1}[S_{V,n}]$ ...


4

This is clear: the components of $f$ are continuous. Now use the universal property of product spaces.


4

I think it is much clearer to stipulate that $U$ is an open set. Expressing it as an element of the topology does not increase the clarity of what you are saying.


0

A set is called open if it is in the topology. It's consequently implied, by virtue of $U$ being called "open", that it's in the collection $\frak{T}$. While technically correct, I think you'd be wasting time by writing it, better to just declare it "open."


2

I think it might raise eyebrows for those who are new to set theory and topology, but that is precisely why I like this notation. One better get used to certain sets having elements which are sets containing further elements themselves. That said, if you want to “translate” Let $U$ be an open set that contains $x$ a better solution in this case ...


1

A function $f:A\to B$ consists of three "ingredients": A domain $A$, a codomain $B$, and a relation $R_f\subseteq A\times B$ satisfying the following properties: For each $a\in A$, there exists $b\in B$ with $(a,b)\in R_f$; For each $a\in A$ and $b,c\in B$, if $(a,b)\in R_f$ and $(a,c)\in R_f$, then $b=c$. We usually identify the function with $R_f$, and ...


4

If $X$ is a set, the identity map $\mathrm{id}_X\colon X\to X$ is defined by $\mathrm{id}_X(x) = x$. Note that the domain and codomain of $\mathrm{id}_X$ are both $X$. If $X$ is a subset of $Y$, the inclusion map $i\colon X\to Y$ is defined by $i(x) = x$. So $i$ is defined by the same formula as $\mathrm{id}_X$, but the codomain is allowed to be different. ...


1

In the notation of your post, we want to show that if the arithmetic sequences $B_1$ and $B_2$ have a point $z$ in common, then they have a whole infinite arithmetic sequence $B_3$ in common. Since $B_1$ and $B_2$ have $z$ in common, there are integers $x_1$ and $x_2$ such that $z=a_1x_1+b_1=a_2x_2+b_2$. Let $B_3$ be the set of all numbers of the shape $...


4

This is false in general if $Y $ is not Hausdorff. Take any map $f : X \to X $, where on the left, $X$ has the discrete topology and on the right the indiscreet topology (only $\emptyset $ and $X$ are open), where $X $ is any finite set. Finally, let $A=\{x\} $. Is is not hard to see that this is a counterexample (if $X $ has more than one element).


5

The claim can be made false without the assumption that $Y$ is Hausdorff. To see this, let $Y\equiv\{1,2\}$ be endowed with the indiscrete topology (that is, the only open sets are the empty set and the whole set). Let $X\equiv\{1,2\}$ be endowed with the discrete topology (all subsets are open) and $f(1)\equiv 1$ and $f(2)\equiv 2$. Clearly, $f$ is ...


4

As the other answers indicate, compactness and connectedness have little to do with one another. You may be thinking in terms of compact cubes, so you will have to think of more general sets. For example, $[0,1]\cup[2,3]$ is compact but not connected, and is made up of compact intervals.


1

As a sort of repeat to what the others have said: in $\mathbb{R}^n$, compactness is equivalent to closed and bounded (Heine-Borel theorem). So, with that in mind, it should be easy to construct counter-examples.


3

No. For example, a set consisting of two points is not connected but is compact.


5

Finite sets are compact, and never connected unless they have one point (or none). The Cantor set is disconnected (totally disconnected even), or more simply: take two disjoint compact sets and take their union: this is still compact but always disconnected. Etc. So there is no relation.


2

Hint:Consider $\{A_n^c\}_{n=1}^{\infty }$ is the open covering of $V$.


1

Hint: Consider the set $\mathcal{U}:=\{U_n |\ n \in \Bbb{N}, U_n:=\Bbb{R}^d\setminus A_n\}$. Each member of $\mathcal{U}$ is open since $A_n$ is closed for each $n$. Furthermore, since $V\cap (\bigcap_\Bbb{N} A_n) = \emptyset$, we have $$V\subset \Bbb{R}^d\setminus \bigcap_\Bbb{N} A_n= \bigcup_\Bbb{N} \Bbb{R}^d\setminus A_n = \bigcup_\Bbb{N} U_n.$$ Then $\...


0

Basically, what you want to show is that any decreasing sequence of nonvoid compact sets has a nonvoid intersection. Let $\{F_n\}$ be such a sequence. Pick $x_n\in F_n$ for all $n\in\mathbb{N}$. This is a sequence lying in a compact set. It has a subsequential limit $x$. You can now show $x\in\cap_n F_n.$


3

The pre-order being complete has no topological meaning, but purely set-theoretic. It means that for any two points $x,y$ in the domain of the pre-order, we must have $x \succsim y$ or $y \succsim x$ (or possibly both here, because we have a pre-order, so we can have both at the same time (invariant goods (?), or some such thing, economics is not my field, ...



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