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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