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Jan
5
comment How to prove $C_1 \|x\|_\infty \leq \|x\| \leq C_2 \|x\|_\infty$?
@Fabian: Assuming that $\| x \|_\infty := \max_{i \in \{1, \dots , n\} } x_i$ I think you are missing an $n$ in the following line: $$ \| x - y \| = \dots \leq n \| x - y\|_\infty$$
Jan
4
revised Finding limit for the function
One brace deleted.
Jan
4
revised $C_c(X)$ dense in $L_1(X)$
added 38 characters in body
Jan
3
comment $C_c(X)$ dense in $L_1(X)$
Dear @MattE, thanks for pointing this out. I think it's fixed now.
Jan
3
revised $C_c(X)$ dense in $L_1(X)$
Different version of Tietze added as discussed with tb to fix mistake pointed out by MattE.
Jan
3
answered $C_c(X)$ dense in $L_1(X)$
Jan
3
comment Picard's method application
I assume you meant to write "on which interval".
Jan
3
answered Smooth functions with compact support are dense in $L^1$
Jan
2
revised Proof of $f \in C_C(X)$ where $X$ is a metric space implies $f$ is uniformly continuous
added 64 characters in body
Jan
2
comment Proof of $f \in C_C(X)$ where $X$ is a metric space implies $f$ is uniformly continuous
@Johan For a continuous function $f$ I have that for an $\varepsilon$ I can find a $\delta$ such that $f(B(x, \delta)) \subset B(f(x), \varepsilon)$. Above I have chosen $\delta_i$ such that $f(B(x_i, \delta_i)) \subset B(f(x_i), \varepsilon)$. I also have $x,y \in B(x_i,\delta_i)$ so $f(x), f(y) \in f(B(x_i,\delta_i)) \subset B(f(x_i), \varepsilon)$ and therefore $|f(x) - f(y)| < 2 \varepsilon$. Can you point out my thinko more explicitly please? I still don't see it, thank you.
Jan
2
revised Proof of $f \in C_C(X)$ where $X$ is a metric space implies $f$ is uniformly continuous
added 4 characters in body
Jan
2
accepted Proof of $f \in C_C(X)$ where $X$ is a metric space implies $f$ is uniformly continuous
Jan
2
comment Proof of $f \in C_C(X)$ where $X$ is a metric space implies $f$ is uniformly continuous
@t.b. That was a "thinko", not a typo. Yes, of course it's less than $2 \varepsilon$. Thank you!
Jan
2
asked Proof of $f \in C_C(X)$ where $X$ is a metric space implies $f$ is uniformly continuous
Jan
1
awarded  Disciplined
Jan
1
accepted $H_1(X,A) = 0 \iff H_1(A) \rightarrow H_1(X)$ surjective and $X_i$ contains no more than one path-component of $A$
Jan
1
accepted $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian
Jan
1
accepted Equivalent identification to get the projective plane?
Jan
1
accepted Homology of disjoint union is direct sum of homologies
Jan
1
accepted Proof of another Hatcher exercise: homotopy equivalence induces bijection (part II)