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May
28
revised Determining if $\int f_n\to 0$ implies that $f_n\to 0$ in measure and $f_n(x)\to 0$ a.e.
added 21 characters in body
May
28
revised Determining if $\int f_n\to 0$ implies that $f_n\to 0$ in measure and $f_n(x)\to 0$ a.e.
added 15 characters in body
May
28
revised Determining if $\int f_n\to 0$ implies that $f_n\to 0$ in measure and $f_n(x)\to 0$ a.e.
added 4 characters in body
May
28
answered Determining if $\int f_n\to 0$ implies that $f_n\to 0$ in measure and $f_n(x)\to 0$ a.e.
May
13
comment Let $(X, \mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then 1. $Cl(A) = Cl(Int(A))$ 2. $Int(A) = Int(Cl(A))$
@AlyssaWallace: No problem. I'm glad you found it helpful.
May
13
comment Let $(X, \mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then 1. $Cl(A) = Cl(Int(A))$ 2. $Int(A) = Int(Cl(A))$
@AlyssaWallace. Sure. You're welcome
May
13
comment Let $(X, \mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then 1. $Cl(A) = Cl(Int(A))$ 2. $Int(A) = Int(Cl(A))$
@AlyssaWallace: The whole real line is an open set so its interior is the whole real line itself.
May
12
comment boundary of $\mathbb{Q}^n$ in $\mathbb{R}^n$
$\partial\mathbb{Q}$ is not $\mathbb{Q}$.
May
12
answered Let $(X, \mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then 1. $Cl(A) = Cl(Int(A))$ 2. $Int(A) = Int(Cl(A))$
May
12
comment Prove that $X$ contains exactly two clopen sets if and only if every nonempty proper subset of $X$ has a nonempty boundary.
This looks correct. In the first direction you probably don't mean that it's closure is empty but boundary. Maybe a typo.
May
12
revised Sum of open/closed/compact sets in $\mathbb{R}^n$ open/closed/compact
added 40 characters in body
May
12
comment Sum of open/closed/compact sets in $\mathbb{R}^n$ open/closed/compact
@JackLee: Right, good call. I totally overlooked that one. Thanks for pointing it out, I'll edit it.
May
12
answered Sum of open/closed/compact sets in $\mathbb{R}^n$ open/closed/compact
May
11
answered $[0,1]\times\mathbb{N}/(0,k)$ not metrizable.
May
11
comment injective/path component
@smith: So you have concluded that $\pi_0(\{0,1\})$ is a $2$-point set, and $\pi_0([0,1])$ is a $1$-point set. Now, there is only $1$ map from a $2$-point set to a $1$-point set. What is it? Is it injective?
May
11
comment injective/path component
@smith: Once you know what $\pi_0(\{0,1\})$ and $\pi_0([0,1])$ are in this particular case, you will realize that there will only exist one map from the first to the latter. This must precisely be the induced map: you don't even need to calculate it explicitly.
May
11
comment injective/path component
@smith: Exactly. So what are $\pi_0(\{0,1\})$ and $\pi_0([0,1])$? Now note that any map defined on a discrete set is continuous. The only injective ones is the identity map and the permutation map. Take the identity map for example. What must the induced map $f_*$ be? And why?
May
11
comment injective/path component
@smith: What are the clopen sets of the discrete $2$-point set? What about the interval $[0,1]$?
May
11
comment injective/path component
@smith: The unit interval is path connected within the reals.
May
11
answered injective/path component