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7h
comment show that the function $\{x_n\}\mapsto \sum_{n=1}^\infty 2^{-n}x_n$ is continuous
@Rubertos If you mean "metrizable by $d(\{x_n\},\{y_n\}) = \sum 2^{-n}\lvert x_n - y_n\rvert$", then yes, that is fine. Showing that that metric induces the product topology is however stronger than showing the continuity of $f$ here. But sooner or later one should prove that such a construction induces the product topology on a product of countably many metric spaces. See here.
7h
comment show that the function $\{x_n\}\mapsto \sum_{n=1}^\infty 2^{-n}x_n$ is continuous
Much easier: $\{ x_n\} \mapsto x_k$ is continuous for every $k$. Uniform convergence.
10h
revised GCD of many numbers divisible by another number
edited tags
10h
revised How did this GCD formulae came about
edited tags
10h
revised What is the maximum value of the LCM of three numbers $\leq n$, as a function of $n$?
edited tags; edited tags
12h
comment Find integer solution of sysem of quadratic equations
Start with adding the two equations.
13h
comment Graph of a non-continuous function is closed
The lemma only gives you the closedness of $\Gamma_{f\lvert_{X_k}}$ in $X_k \times \mathbb{R}$, not in $\mathbb{R}\times \mathbb{R}$. For $a \leqslant 0$, that's no problem, since $(-\infty,0]$ is closed in $\mathbb{R}$, but for $a > 0$, it is a problem.
13h
comment A closed set is not a submanifold
The boundary points - $x = 0$ - don't fit the definition.
13h
comment Show that $\|f\|^{2}$ attains a minimum value on the interior of $B$
@user135520 I've elaborated. If anything is unclear, don't hesitate to ask.
13h
revised Show that $\|f\|^{2}$ attains a minimum value on the interior of $B$
Expand the argument
14h
comment Show that $\|f\|^{2}$ attains a minimum value on the interior of $B$
@user135520 Sure. But first, can you confirm that $B$ is the closed unit ball, or is it the open unit ball? [For the closed unit ball, we have a simpler way to give the proof by using the boundary sphere, but it's not a fundamental difference.]
16h
comment Inequality of logarithm of the tail of the Euler product
Since $\lim\limits_{b\downarrow 1} \sum\limits_{p > x} \log (1-p^{-b}) = -\infty$, such an inequality cannot hold for all $b > 1$. For every $x$, there is a $b_x > 1$ such that the inequality fails for $b < b_x$.
16h
comment Inequality of logarithm of the tail of the Euler product
It's false for $b = 1$, then the left hand side is $-\infty$. Can we ignore that case and assume $b \geqslant 2$?
20h
comment Sum of the reciprocals topology
It's not quasicompact. $F_k = \{ 2^n : n \geqslant k\}$ is closed for all $k$, the family has the finite intersection property, but empty intersection.
20h
comment Is the direct sum of two orthogonal subspaces well defined in infinite-dimensional vector spaces?
Literature recommendations are always a weak point for me. You would need to read (at least parts of) some book on Functional Analysis, but I'm not sure what might be a good choice for you. If you like Rudin's style, his "Functional Analysis" is of course a worthwhile read. But there are lots of good books on FA, I suggest you look at a couple and pick one whose style suits you.
21h
comment Is the direct sum of two orthogonal subspaces well defined in infinite-dimensional vector spaces?
I have given a proof of the projection lemma here.
21h
comment Is the direct sum of two orthogonal subspaces well defined in infinite-dimensional vector spaces?
No, if you consider $\ell^2(\mathbb{N})$, the proper subspace of the sequences with only finitely many nonzero terms is dense, so it's not closed. It's not very difficult to prove (the part that you have a direct sum $M\oplus M^\perp$ is very easy, that that direct sum is $V$ if $M$ is closed and $V$ complete is a bit of work). Once you have the projection lemma, the rest is immediate.
21h
comment $\lim_{x\to{\infty}} (x-\sqrt{x^2+x})$
One of the easiest ways.
21h
comment Is the direct sum of two orthogonal subspaces well defined in infinite-dimensional vector spaces?
The direct sum $M \oplus M^\perp$ is well-defined. It need not be equal to $V$, though. If $V$ is complete and $M$ is closed, then you have $V = M \oplus M^\perp$.
21h
comment Regularization of a (divergent) cosine series
Related, though I'm not sure I'd call it a duplicate.