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9

Yes, consider $X := \Bbb Z$ endowed with the discrete topology. For any topological manifold $M$, $\dim (M \times M) = \dim M + \dim M = 2 \dim M$. Since the dimension of a nonempty topological manifold is well-defined, there is no positive-dimensional topological manifold $M$ for which $M \cong M \times M$, which in particular excludes $R$ and $S^1$ as ...


5

Take $$ X = [0,1] \cup [2, \infty), $$ under (the usual) order topology. If $a = 0, b = 2$, then we get $$ \overline{(a,b)} = [0,1]. $$ On the other hand, $$ [a,b] = [0,1] \cup \{2\}. $$


4

I am not entirely sure what your question is, but here is my interpretation of it: is every $n$-manifold $X$ (without boundary) the boundary of some $(n+1)$-manifold with boundary $Y$? The answer is yes: just take $Y=X\times [0,\infty)$, identifying $X$ with $\partial Y=X\times\{0\}$. (Mike Miller gave an answer to the contrary in the comments; however, ...


4

Many Banach spaces are linearly homeomorphic to their Cartesian squares. For instance all classical spaces including $c_0$, $\ell_\infty$, $C(K)$ for $K$ compact metric, $L_p(\mu)$ for $p\in [1,\infty]$ etc.


3

At this level of generality you can make $X=X \times X$ happen quite easily. Take a discrete space of any infinite cardinality, for instance. Or topologize $X=A^B$ by whatever means and compare $X \times X = A^{B \sqcup B}$; under various mild assumptions on $B$ those spaces would be homeomorphic.


2

Hint. You can use Banach fixed point theorem for the operator $T$ restricted to the space $\mathcal C([0,1-\frac{1}{n}])$ for $n \ge 2$ as $T$ is a contraction on this space. For each $n$ you get a unique continuous function $f_n$ defined on $[0,1-\frac{1}{n}]$ solution of the Banach fixed point problem (the one defined by $T$ operator). By unicity, if $n ...


2

Take $a=b$. Then $\overline{(a,a)} = \overline\emptyset = \emptyset$, while $[a,a] = \{a\}$.


2

Follow your nose: $|x_i + y_i| \leq |x_i| + |y_i| \leq \sup_i|x_i| + \sup_i|y_i| \quad \forall\,i,$ take the supremum and then conclude. I find it very weird using $\scr L^2$ here though - one would use $\scr L^\infty$ for this space, and $\scr L^2$ for the sequences with $\sum_i |x_i|^2 < +\infty$ (or even $\ell^\infty$ and $\ell^2$).


2

Note that \begin{align*} \underbrace{\color{red}{|a_1-c_1|}+\color{red}{|c_1-b_1|}}_{\clubsuit}\leq&\,\underbrace{\max\{\color{red}{|a_1-c_1|},|a_2-c_2|\}+\max\{\color{red}{|c_1-b_1|},|c_2-b_2|\}}_{\star},\\ ...


2

You're overthinking this. Let $$\ell^2 = \left\{ \{x_n\} : \sum_{n=0}^\infty |x_n|^2<\infty\right\}. $$ If $x,y\in\ell^2$ then for each $n$, $|x_n-y_n|^2\leqslant 2|x_n|^2 + 2|y_n|^2$, so $$\sum_{n=0}^\infty |x_n-y_n|^2 \leqslant 2\sum_{n=0}^\infty |x_n|^2 + 2\sum_{n=0}^\infty |y_n|^2 <\infty, $$ so that $x-y\in\ell^2$ and ...


2

what is the Significance / usefulness of a set being closed? To this very broad question, I would say that, in my opinion, in many instances where I have seen that establishing the closure of a set is important is because it is a necessary condition for compactness in complete normed spaces. what more can be deduced when a set is proved to be closed ...


1

Well, that's actually pretty easy. Just note that the orders are isomorphic, since they are both complete orders with dense countable subsets and no maximum. There is an isomorphism between the countable dense subsets, and it extends uniquely to the rest of the order. More specifically, you need to pick an order embedding of $\alpha$ into $\Bbb R$, or ...


1

Consider an open ball of radius $r$ around an element $x\in Z$ under the metric $d.$ For $r\leq 2,$ this is precisely a cylinderset, and for $r>2,$ this is $Z.$ Both of these are open under the product topology. Since every open ball under the metric topology is open under the product topology, the product topology also contains the metric topology.



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