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3h
accepted Cocountable topology and topological vector spaces
18h
asked Cocountable topology and topological vector spaces
Jul
19
comment The space of continuous, bounded functions from a metric space $X$ to $\mathbb R$
In your first point you show $\lVert f_n- f \rVert < \frac 2 3 \epsilon$ if $n \geq n_1$, which seems strange to me since $n_1$ dependes on a chosen $x \in X$. As I see it, you cannot simply take the supremum if you don't have a uniform $n_0$ which is valid for all $x \in X$.
Jun
30
asked $K_0$ of a ring via idempotents
Jun
29
comment Direct (inductive) limit of groups
Is a filtered category the same as a directed category ? And have categories which correspond to a directed poset (poset with finite upper bounds) a special name ?
Jun
29
comment Direct (inductive) limit of groups
Yes, everything is functorial and each pair $i,j$ has an upper bound. Furthermore I would write rather $f_{jk} f_{ij} = f_{ik}$, than $f_{ij}f_{jk} = f_{ik}$.
Jun
29
comment Direct (inductive) limit of groups
Which meaning has "directed colimit" ?
Jun
29
accepted Direct (inductive) limit of groups
Jun
29
asked Direct (inductive) limit of groups
Jun
28
reviewed Approve Find $\int_a^b \sin |x| \, \mathrm{d}x $
Jun
26
comment Mathematical Operator to flip a vector
Your matrix is not an operator ?
Jun
25
accepted Finitely generated and free module
Jun
24
comment Finitely generated and free module
How one can prove my assertion, that a free and f.g. module is isomorphic to $R^n$ for some $n$.
Jun
24
asked Finitely generated and free module
Jun
24
accepted Category of torsion free abelian groups not abelian
Jun
24
revised How to find the image of an arbitrary element under this operator?
added 28 characters in body
Jun
24
answered How to find the image of an arbitrary element under this operator?
Jun
24
comment How to find the image of an arbitrary element under this operator?
I don't agree with myself, too. Sorry for the confusion. To elaborate on why you can extend $T$ to $H$: The norm of $T$ on the span of the basis vectors is $1$, hence $T$ can extended to $H$, having norm $1$ on $H$.
Jun
24
comment How to find the image of an arbitrary element under this operator?
For this definition of $T$ to work, you implicitly assume that $T$ is bounded. Otherwise you can not extend $T$ to $H$, when it is only defined for basis vectors.
Jun
23
asked Why not take the tensor product of two left modules in this way?