Let $X$ be a metric space ; then which of the following is possible ?
1) $X$ has exactly $3$ dense subsets
2) $X$ has exactly $4$ dense subsets
3) $X$ has exactly $5$ dense subsets
4) $X$ has exactly $6$ dense subsets
I know that if $X$ has a proper dense subset then for some $a \in X...
Suppose $(X,d)$ is a compact metric space and $f:X\to X$ a continuous map. Show that $f (A)=A$ for some nonempty $A\subseteq X.$
I start this by supposing that $A_0:=X$ and $A_{n+1}:=f(A_n)$ for all $n \geq 0$. If $A_n=A_ {n+1}$ for some $n$ then the purpose is done. But if not, how can we t...
This is based on Ex. 6.4.6 in Stillwell's "Real Numbers."
Using previous exercises, it was established that one can construct for any countable ordinal $\gamma$ disjoint half-open intervals $[a_{\alpha}, a_{\alpha+1})$ for all $\alpha\lt\gamma$, with the properties $a_{\alpha}\lt a_{\beta}$ iff ...
Let $X$ be a metric space ; then which of the following is possible ?
1) $X$ has exactly $3$ dense subsets
2) $X$ has exactly $4$ dense subsets
3) $X$ has exactly $5$ dense subsets
4) $X$ has exactly $6$ dense subsets
I know that if $X$ has a proper dense subset then for some $a \in X...
Suppose $(X,d)$ is a compact metric space and $f:X\to X$ a continuous map. Show that $f (A)=A$ for some nonempty $A\subseteq X.$
I start this by supposing that $A_0:=X$ and $A_{n+1}:=f(A_n)$ for all $n \geq 0$. If $A_n=A_ {n+1}$ for some $n$ then the purpose is done. But if not, how can we t...
