Reputation
1,159
Top tag
Next privilege 2,000 Rep.
Edit questions and answers
Badges
4 26
Impact
~19k people reached

May
22
comment Revised proof for the set of positive irrational numbers closed under multiplication*
I would most definitely advise the poster to read this handout on good mathematical writing: math.hmc.edu/~su/math131/good-math-writing.pdf
May
17
comment Dimension of Vector Spaces
Good, see what properties must hold in specific cases to write a basis.
May
17
comment Dimension of Vector Spaces
Try computing the bases for these spaces. The dimension will simply be the number of vectors in the basis. Hint: For 3, this is the set of products of $\mathbb{R}^m$ and $\mathbb{R}^n$ so that it has dimension $\mathbb{R}^{n+m}$. That is, if $\mathcal{A}_n$ is a basis for $\mathbb{R}^n$ and $\mathcal{A}_m$ is a basis for $\mathbb{R}^m$, then $\mathcal{A} = \mathcal{A}_n \cup \mathcal{A}_m$ is a basis for the space of $m \times n$ matrices.
May
17
comment Dimension of Vector Spaces
do you know the notion of a vector basis?
May
11
asked Necessity of continuity in Topological Vector Space
May
9
accepted No continuous injective map $f: \mathbb{S}^1 \to \mathbb{R}$
May
9
comment No continuous injective map $f: \mathbb{S}^1 \to \mathbb{R}$
@DanielFischer How now, do we put this all together. That is, the notion of $f$ attaining its extrema, and this homeomorphic property with $(-1,1)$?
May
9
comment No continuous injective map $f: \mathbb{S}^1 \to \mathbb{R}$
@DanielFischer Formally?
May
9
comment No continuous injective map $f: \mathbb{S}^1 \to \mathbb{R}$
Are you specifically saying to apply the MVT in this case? I don't see how to do this outside of $\mathbb{R}$.
May
9
comment No continuous injective map $f: \mathbb{S}^1 \to \mathbb{R}$
@DanielFischer What is this property? Again, intuitively, it has something to do with the fact that I stated in my question.
May
9
asked No continuous injective map $f: \mathbb{S}^1 \to \mathbb{R}$
May
5
revised Finite dense subset implies $X$ finite
added 21 characters in body
May
5
asked Finite dense subset implies $X$ finite
May
4
comment Convergence of a sequence by convergence of sub-subsequence
@John Yes, I've done that...it's rather easy. But I'm searching for a direct proof.
May
4
comment Convergence of a sequence by convergence of sub-subsequence
@John How can I remedy this?
May
4
comment Convergence of a sequence by convergence of sub-subsequence
@John Unless I'm mistaken.
May
4
comment Convergence of a sequence by convergence of sub-subsequence
@John I'm not sure that I see why.. if this set were infinite then the subsequences wouldn't converge for $k=N$ large enough. I'm saying what's left over is finite, $B$.
May
4
comment Convergence of a sequence by convergence of sub-subsequence
@John Yes, we're allowed to assume this is the same $p$.
May
4
asked Convergence of a sequence by convergence of sub-subsequence
Apr
16
accepted Integral equal to Riemann Zeta Function