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Jun
4
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
@tomasz What's the difference between direct sum and "simple internal sum"? Doesn't $X=A \oplus B$ imply $X = A + B$ if $A,B$ are subrings of some $X$?
Jun
4
comment A question on countability of isolated points of a subset of R
@Seirios Oops, you are right. Thank you for your reply.
Jun
3
comment A question on countability of isolated points of a subset of R
I was wondering why you chose to use two rational points. Couldn't one argue that since $x$ is isolated there is an $\varepsilon$-ball such that $B(x,\varepsilon) \cap A = \{x\}$. Then there exists a rational $q$ in this ball. Define $\phi(x) = q$.
Jun
2
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
Thank you for your comments. I will get back to you, right now (for the next few hours) something else is keeping me from thinking about this.
Jun
2
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
In you answer is $R$ a commutative ring or an algebra over a field? I assume you assume $\phi: R\to K$ for some field $K$. I don't want to bother you, it's just really not 100% clear to me.
Jun
2
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
Exactly. Me neither. : )
Jun
2
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
@seaturtles Yes. But in my previous comment $a$ was an element of the algebra not the the underlying field : )
Jun
2
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
@seaturtles I read your previous comment as "given any algebra homo. $\phi : A \to B$ then $\phi(a) = \phi(a1) = a \phi(1) = a$". Of course, this doesn't even make sense unless $A \subseteq B$.
Jun
2
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
Wait, let me read the edited comment.
Jun
2
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
I see you edited your previous comment.
Jun
2
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
@seaturtles I kept re-reading this to see the mistake but there seems to be none. What is puzzling me is: wouldn't this mean that every algebra homomorphism is the identity?
Jun
2
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
@seaturtles Thank you for your comment. But it is not clear to me why the inclusion map is an inverse of $\phi$.
Jun
1
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
@tomasz Great, thanks for your comment, I misinterpreted the question.
Jun
1
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
I don't know the answer off the top of my head and right now I don't have time to think. I'll get back to you.
Jun
1
comment Proof that if f is function, continuous on an interval I then f(I) is also an interval
Just a minor thing: If you have $f(\alpha)<\lambda< f(\beta)$ you can't have $\lambda=f(\alpha)=f(\beta)$. You'd have to slightly rephrase this. And I'd change $x_\lambda \in I$ into $x_\lambda \in (a,b)$ just to be more precise. But it's not necessary. Otherwise this looks good to me!
Jun
1
comment Proof that if f is function, continuous on an interval I then f(I) is also an interval
@Bardo That's great. If you like you can post it then people on this site can check it for you.
Jun
1
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
One sufficient condition would be the conditions necessary to apply the splitting lemma. The splitting lemma tells you that if you have a short exact sequence: $$ 0 \to \mathrm{Ker}(\phi) \xrightarrow{i} A \xrightarrow{\phi} K \to 0$$ then $A$ is isomorphic to the direct sum of $K$ and $\mathrm{Ker}(\phi)$ if there exists an algebra homomorphism $t:K \to A$ such that $\phi \circ t$ is the identity on $K$.
Jun
1
comment Proof that if f is function, continuous on an interval I then f(I) is also an interval
@Bardo No, I agree with you, as stated it doesn't make much sense. But it's clear to me what it's trying to do: it wants to show that $f(I)$ is an interval by applying the intermediate value theorem. It completely fails to mention what lambda is and that is must be between $\alpha$ and $\beta$. On top of that, the English is broken. Maybe it would be helpful to you if (in addition) you could get yourself a better book from the library.
Jun
1
comment Please verify this definition of locally convex vector space
@luka5z If you found this answer helpful then you might consider upvoting it (in addition to accepting).
Jun
1
comment Please verify this definition of locally convex vector space
I think so. ${}$