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Jul
9
comment coproduct of lattices preserving filtered property of positive elements
In this construction then both meets and joins are computed component-wise. But then $(0,y)\wedge (x,0)=(0,0)$ for all $x\in L_1$ and $y\in L_2$. So if $z\in W$ is well above $0$, then either $z\ge (x,0)$ or $z\ge (0,y)$, with $x\succ 0$ and $y\succ 0$. But these are not closed under finite meets, so $W$ is not filtered.
Jul
9
comment Why do left and right switch when direction is reversed?
@ErickWong, yes, once that 180 degree turn is taken in order to trace your steps back. It is no longer a turn, but rather a reflection. Locally, you do the same, but globally things are very different.
Jul
9
answered Why do left and right switch when direction is reversed?
Jul
8
comment coproduct of lattices preserving filtered property of positive elements
@JonMarkPerry thanks, and corrected.
Jul
8
revised coproduct of lattices preserving filtered property of positive elements
edited body
Jul
7
answered If $G$ is abelian, it has a subgroup in every order of $|G|'s$ divisors?
Jun
30
comment Convergent $x_n,y_n$ and $x_n^{y_n}$ diverges
To dispel any illusion that the sought counterexample is hard it should be noted that one can construct trivial counterexamples easily. While this example is pretty, it somewhat misses the point of the exercise.
Jun
30
comment Convergent $x_n,y_n$ and $x_n^{y_n}$ diverges
the problem is that the quantity $0^0$ is not well-defined. The book is fine.
Jun
30
answered Convergent $x_n,y_n$ and $x_n^{y_n}$ diverges
Jun
28
answered In the Hahn-Banach theorem, what is the purpose of the 'dominating function'?
Jun
28
asked coproduct of lattices preserving filtered property of positive elements
Jun
25
answered Is the Cauchy-Schwarz inequality ever used in Physics?
Jun
25
comment Prove that the image of a a closed and bounded interval in $\mathbb{R}$ is a a closed and bounded interval in $\mathbb{R}$?
You can't use compactness to prove the result. The image of a closed interval will be compact but there are lots of compact subsets of $\mathbb R $ which are not closed intervals.
Jun
23
comment Duality of Projective and Inductive Limit
In the sense that projective limit in $C^op$ is the same as injective limit in $C$. This is the meaning of duality in category theory.
Jun
23
answered Cauchy sequences of two metric spaces with only different distance functions
Jun
23
comment Countable Connected Hausdorff Space
did you google "countable connected Hausdorff space"? You will get lots of hits.
Jun
23
comment $\mathbb{Q}(\sqrt2) $ is isomorphic to $\mathbb{Q}(\sqrt2 +2 )$
If they are equal, then they are isomorphic. You don't need any homomorphism (and I'm not sure which homomorphism you had in mind. The one you tried to describe is unclear.)
Jun
23
comment $\mathbb{Q}(\sqrt2) $ is isomorphic to $\mathbb{Q}(\sqrt2 +2 )$
They are equal, so yes, they are isomorphic.
Jun
21
comment How should one picture a topology/ topological space?
no, still not a topology and the last paragraph is still not good.
Jun
21
comment How should one picture a topology/ topological space?
the collection of all intervals in $\mathbb R$ is not a topology. You last paragraph is obscure. What do you mean by "take some arbitrary set, draw open balls lying in..." how can you draw open balls in an arbitrary set? what does that mean? The rest is just as unclear.