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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
May
10
comment Path components
@smith: Yes, it won't be connected. But the two components are path components. Note that $f_+$ and $f_-$ are continuous functions and the sets are continuous images of path connected sets.
May
10
revised Equivalence of sequence spaces
edited tags
May
10
answered Equivalence of sequence spaces
May
10
answered Path components
May
10
revised Example 3, Sec. 25 in Munkres' TOPOLOGY, 2nd ed: Why is the topologist's sine curve not locally connected?
added 18 characters in body
May
10
comment Example 3, Sec. 25 in Munkres' TOPOLOGY, 2nd ed: Why is the topologist's sine curve not locally connected?
@SaaqibMahmuud: For each $k$, we have $\emptyset\neq A_k\neq I_k$, so $I$ is just a countable union of disjoint disconnected sets.
May
10
awarded  Cleanup
May
10
revised Example 3, Sec. 25 in Munkres' TOPOLOGY, 2nd ed: Why is the topologist's sine curve not locally connected?
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