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Feb
6
comment what is non trivial basis for cofinite topology on non empty set $X$
I just noticed: In order to obtain a basis, we also need the condition $\bigcap\mathcal{F}=\varnothing$. Otherwise, take $x\in\bigcap\mathcal{F}$. Then there exists no $F\in\mathcal{F}$ with $x\in X\setminus\mathcal{F}$.
Feb
6
comment what is non trivial basis for cofinite topology on non empty set $X$
To see that all bases are of the form you described: If $\cal{B}$ is a basis, let $\mathcal{F}=\left\{F:X\setminus F\in\cal{B}\right\}$. Then clearly $\cal{B}=\cal{B}_{\cal{F}}$. To check that $\cal{F}$ is cofinal, let $G$ be any finite subset of $X$. Then there exists $B\in \cal{B}$ with $B\subseteq X\setminus G$, so $G\subseteq X\setminus B\in\cal{F}$.
Jan
18
awarded  Yearling
Nov
24
awarded  real-analysis
Nov
19
comment Proving that a one-to-one continuous function on a compact subset has a continuous inverse
More or less. Take any closed $D\subseteq A$. Then we need to verify that the preimage of $D$ under $f^{-1}$ is closed. This preimage is precisely $(f^{-1})^{-1}(D)=f(D)$. I made a mistake in my first comments, and should have writte "$f(D)$ is compact" in the end. Anyway, we've shown that preimages of closed sets under $f^{-1}$ are closed, so $f^{-1}$ is continuous. This is the definition of homeomorphisms.
Nov
19
answered Commutative free products
Nov
19
comment Proving that a one-to-one continuous function on a compact subset has a continuous inverse
Yes, that's one of the several equivalent definitions.
Nov
19
comment Proving that a one-to-one continuous function on a compact subset has a continuous inverse
You probably know that a map $f$ is continuous iff the preimage of open sets is open. Taking complements, this is equivalent to the preimage of closed sets being closed. Now consdering $f^{-1}$, the preimages are the direct images of $f$. If $D\subseteq A$ is closed, it is a closed inside a compact, hence compact, so $f(A)$ is compact, hence closed.
Nov
19
revised Show that the function is not Lebesgue Integrable
added 594 characters in body
Nov
19
answered Show that the function is not Lebesgue Integrable
Nov
9
comment The closure of a connected set is connected
@dustin This was a typo. I meant $U\cap\overline{B}=\varnothing$. Also, $U$ is a neighbourhood of $a$ in $X$, so it does not need to be true that $U\subset A$.
Nov
9
comment The closure of a connected set is connected
@Babai This was a typo. I meant $U\cap\overline{B}=\varnothing$.
Nov
9
revised The closure of a connected set is connected
edited body
Oct
15
comment Prove or disprove the following statement about natural numbers
@ndrizza Yes. Another way of writing it would be $\alpha=(n_0)!+n_0$.
Oct
15
answered How to show two sets are either open or closed in $\mathbb{Q}$?
Oct
15
answered Prove divisibility: if $a\mid (b-d)$ and $a\mid (c-e)$, then $a\mid (bc-de)$
Oct
15
answered Prove or disprove the following statement about natural numbers
Oct
15
revised What are the elements of $2^A$ if $A$ is a set
added 1 character in body
Oct
15
answered What are the elements of $2^A$ if $A$ is a set
Oct
15
answered Limit of a sequence and its infimum