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3

Yes, that is correct. In topology, an open map is a function between two topological spaces which maps open sets to open sets. That is, a function $f : X \rightarrow Y$ is open if for any open set $U$ in $X$, the image $f(U)$ is open in $Y$. Likewise, a closed map is a function which maps closed sets to closed sets. A map may be open, closed, both, or ...


3

It’s false in general. Let $n=2$ and $A=\{0,1\}$. Let $X_0^{(1)}=X_1^{(2)}=\{0\}$ and $X_0^{(2)}=X_1^{(1)}=\{1\}$. Then $$\bigcap_{i=1}^2\bigcup_{\alpha\in A}X_\alpha^{(i)}=\left(X_0^{(1)}\cup X_1^{(1)}\right)\cap\left(X_0^{(2)}\cup X_1^{(2)}\right)=\{0,1\}\cap\{0,1\}=\{0,1\}\;,$$ but $$\bigcup_{\alpha\in ...


3

For a) I would just consider a constant function, $[a,a]$ is a closed interval. If you want an interval of positive length, $\sin$ or some variation (like yours) is the way to go. Personally I would prefer to adapt the domain to adding stuff and factors. For b) your argument is correct for bounded intervals. For c) you could do something with ...


2

$\require{AMScd}$I'd rather think about the negation of your first condition. So call a space bad if there exists a decomposition $A \cup B$ into connected open sets with disconnected intersection. Call a space semi-bad if there exists a decomposition into open sets $A \cup B$ such that the induced map on relative homology $\tilde H_0(A \cap B) \to \tilde ...


2

Most certainly. One interesting way is to write the category of spaces as the category of relational algebras for the ultrafilter monad on sets. This is a fancy way of saying that a space is the data of a set together with the choice of a set of limit points for every ultrafilter on that set, such that principal ultrafilters converge to the point they're ...


2

Well, functional analysis provides very natural set up for dealing with this question. We will prove the same for $A_{c(x)}:= \left \{p(x) e^{-c(x)} \right\}$, where $c(x)$ is a fixed even degree polynomial with positive leading coefficient and of degree $\ge 2$ and $p(x)$ is any polynomial. Note that we need to prove that $A_{c(x)}$ is dense in $C_0 ...


1

$I\times I$ is not convex in $\Bbb R\times \Bbb R$ under the order topology. For example, $(\frac12,42)$ is between $(0,0)$ and $(1,0)$.


1

Let $X$ be a topological space and $S= \varnothing$. A $x\in X$ is a limit point of $S$ if for every $\epsilon >0$ it holds: $$\big(B(x,\epsilon)\setminus \{x\}\big) \cap S \neq \varnothing.$$ But $S = \varnothing,$ thus the above intersection will always be the empty set for any $x\in X$. Hence, there are no $x \in X$ that are limit points of $S$.


1

A counter example to make the statement untrue would be X = the irrationals and Y= the rationals. Then X' = Y' = the reals and $X' \cap Y' = \mathbb R $. But $X \cap Y =\emptyset $. So what is $\emptyset'$? Well, every neighborhood of a limit point of the empty set must have a point of the empty set. But the are no such points to be had so no limit ...


1

The question is a little vague, but I'll do my best to answer. $\mathbf{Top}$ is a concrete category, that is a category with a faithful functor to the category of sets. Here, this faithful functor is the forgetful functor $\mathbf{Top} \to \mathbf{Set}$, which sends a topological space $(X, \tau)$ to its underlying set $X$. Concrete categories can be ...


1

Yes it is. Consider, for example, the continuous function $$f(x) = x^2$$ What is the image of the open set $(-1,1)$ ?


1

Continuous maps don't have to map open sets to open sets. An example is the map $f:\mathbb{R}\rightarrow\mathbb{R}$ given by $f(x)=x^2$ which maps $(-1,1)$ to $[0,1)$ which is not open and not closed.


1

Yes. I tend to think of this (quickly) as "continuous = open sets come from open sets" and more slowly as "the preimage of an open set is an open set". There is another term: A map is open if it takes open sets to open sets. A map may be open, continuous, neither, or both. This is the other idea that a person first learning this definition of continuous ...


1

This looks like mostly good work. a) This is perfect as is. b) Just excluding $\mathbb{R}$ is not enough since you can map $[0,\infty)$ to an open interval. But your argument is valid for bounded intervals. c) If you want to find a map on a finite interval consider the map $(0,1)\to\mathbb{R}$ defined by $x\mapsto \frac{1}{x}$. This doesn't completely ...


1

For b) think about $(\sin x)(1-1/x)$ on $[1,\infty).$


1

This is not the case. Let $X=\{1,2\}$ and $Y=\{1\}$ under the discrete topology. Let $U_i=Y$ and $W_i=\{i\}$ for $i\in\{1,2\}$. Clearly the collection $\{f_1,f_2\}$ satisfies the given conditions, and each $f_i\colon W_i\rightarrow Y$ is a homeomorphism, but $f^{-1}(Y\cap Y)=f^{-1}(Y)=\{1,2\}\neq W_1\cap W_2=\emptyset$.


1

Consider two spaces $A,B$ with equivalence relations $R,S$. Then $R\times S$ is an equivalence relation, and there is a natural map $p:(A\times B)/(R\times S)\to A/R\times B/S$. This is always a continuous bijection, but might not be an homeomorphism. In your case since the spaces are compact Hausdorff, it is. One can weaken the hypothesis, however. I ...



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