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1d
accepted Confusion between an element and its preimage
1d
comment The Coproduct of two spaces is the same as the disjoint union and is homeomorphic to the union when the spaces are disjoint
In particular what topology could we put on $[0,1)\cup \{1\}=[0,1]$ that could make $[0,1)\cup \{1\}=[0,1]\cong [0,1)\sqcup \{1\}$ and if this homeomorphism exists then $[0,1]$ would be disconnected !
1d
asked Confusion between an element and its preimage
1d
comment The Coproduct of two spaces is the same as the disjoint union and is homeomorphic to the union when the spaces are disjoint
Ok thanks alot.. so in this post math.stackexchange.com/questions/554333/… they are able to find counterexamples by choosing to put the subspace topology on the set $X\cup Y$ instead of the one you mentioned?
1d
comment The Coproduct of two spaces is the same as the disjoint union and is homeomorphic to the union when the spaces are disjoint
Please drhab what do you mean by a sortlike topology ?
1d
asked The Coproduct of two spaces is the same as the disjoint union and is homeomorphic to the union when the spaces are disjoint
May
6
accepted Any square matrix is equivalent to zero diagonal matrix
May
4
asked Any square matrix is equivalent to zero diagonal matrix
May
4
awarded  Yearling
Apr
20
revised Decomposition of the image of a projection.
added 12 characters in body
Apr
20
comment Decomposition of the image of a projection.
Dear Marc it is actually $p\circ q=0$ instead of $q\circ p=0$ but i see what you are doing, the only thing i don't see is the last step how did you go from $q((1-p)(w_0))$ to $q((1-p)(v_1+w_0))$?
Mar
29
awarded  Popular Question
Mar
25
comment Limit of the absolute value of a function
@C-S yes i reread it. Thanks !!
Mar
23
comment Needing a double inclusion to determine the kernel of a matrix
So you agree that we need the other inclusion and what we have by that computation is just the inclusion of the kernel in the span and it is not an equality ?
Mar
23
comment Needing a double inclusion to determine the kernel of a matrix
@AlexHalm If we know that dimension of $\ker f$ is one then it's ok we don't need the other inclusion but here we don't seem to know the dimension of $\ker f$, aren't we ?
Mar
23
asked Needing a double inclusion to determine the kernel of a matrix
Mar
23
comment Limit of the absolute value of a function
I think this is not true take $f(n)=(-1)^n$. While $\lim_{n\to \infty}|f(n)|=1$ but $f(n)$ does not have a limit at infinity.
Mar
22
awarded  Popular Question
Mar
11
awarded  Popular Question
Mar
11
awarded  Popular Question