One can prove directly that it is separable, which is always a bit of work. A more powerful (but elegant) way to prove it may be the following: Note that, using a decomposition of absolutely ...

First, if the map is not surjective, how can you be sure to be able to even say $p^{-1}(\{y\}) \subseteq U_{\alpha}$? You are considering as a trivial step the fact that we can find $W_y$ such that $p^... View answer Accepted answer 2 votes You see that the sequence of functions you have written is$f_{n}(x)= \tanh(nx)$. If you take the limit for$n \rightarrow \infty$you can see that it converges pointwise to a discontinuous function, ... View answer Accepted answer 2 votes The procedure is fine. Just be careful with the substitution in line 4, it is not necessary, but it is not wrong to do it. Nevertheless, you make a mistake since you forget to divide by 2. Another way ... View answer Accepted answer 2 votes A sequence is a function from the natural numbers to the real line, that is you have$a_n: \mathbb{N} \rightarrow \mathbb{R}$, where the pedex$n$is used as$x$in the standard$f(x)$notation. This ... View answer Accepted answer 1 votes You are right. (Formally) the uniqueness of the limit (in the single space you are considering) comes from the fact that normed spaces are$T_2$. But, the main argument in both cases is the ... View answer 1 votes Nope, since every open set in$\mathbb{R}$is made by open intervals (basis for the euclidean topology) and there is no way to intersect an open set of$\mathbb{R}$with$[-5,5]$such that$[0,5]$is ... View answer Accepted answer 1 votes I suppose that for "taking values" you mean the domain of the function, otherwise the notion is not well defined in general (you may solve the problem defining orders on vectors in the sets$...

It is perfectly fine. Another way to prove it (less elegant though) is to assume by contradiction it is connected, then since it is in $A \cup B$ and $A \cap B = \emptyset$ then it must be in only one ...

You need to be careful in writing the domain of the integral and in explaining all the variables. If the question is to show a function is integrable over a domain, yes, you should always check it is ...

The proof is fine and you are right, you don't need $F$ metric space. It holds in general that if $X, Y$ are topological spaces, then $$f: X \rightarrow Y \text{ is continuous }$$ if and only if for ...

First of all a bit of notation: a space cannot be a linear map. A tensor product is a vector space, hence you can consider linear maps having it as the domain. Do not confuse maps having it as domain, ...

Points in $\mathbb{R}^2$ with the euclidean topology are closed sets (every $T_1$ space satisfies this claim). The complement of a closed set is open by definition.