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Consider the following topologies on $X =\Bbb R^n$

  1. The Tychonoff product topology
  2. The box topology
  3. The uniform metric topology

For each of the topologies determine whether $X$ is first countable, second countable, seperable or Fréchet (recall $X$ is Fréchet if whenever $x \in \operatorname{cl}A$ there is a sequence from $A$ converging to $x$)

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u dnt lke vocls, rght? –  gnometorule Feb 5 '13 at 23:18
    
@gnometorule: Eh? –  Asaf Karagila Feb 5 '13 at 23:18
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@Asaf: U mn @gnmtrul. –  Brian M. Scott Feb 5 '13 at 23:20
    
@Asaf Karagila: pre-BMS edit, this read about "Twisty problm: help me getout ths." My comment makes little sense now. –  gnometorule Feb 5 '13 at 23:20
    
He probably means $\mathbb{R}^\mathbb{N}$ instead. –  Henno Brandsma Feb 6 '13 at 10:46
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2 Answers

HINT: You should have no trouble showing that the first two topologies are the same. With just a little more work you can show that the third is the same as the first two. The last one is a metrizable topology, and you can show that $\Bbb Q^n$ is dense, so ... ? (In fact you can show that all of them are just the usual Euclidean topology, whose properties you already know.)

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i need some hints could u explain it little bit more this is not homework problem but could be appear in test –  math Feb 6 '13 at 2:33
    
@motu: You should know a base for each of those topologies, since each of them is usually defined in terms of a base. If $\mathscr{B}_1$ is a base for a topology $\tau_1$, and $\mathscr{B}_2$ is a base for a topology $\tau_2$, one way to prove that $\tau_1\subseteq\tau_2$ is to show that if $x\in B_1\in\mathscr{B}_1$, then there is a $B_2\in\mathscr{B}_2$ such that $x\in B_2\subseteq B_1$. (You should try to see why this is true: it’s fundamental to understanding bases and topologies.) To show that $\tau_1=\tau_2$, show that $\tau_1\subseteq\tau_2$ and $\tau_2\subseteq\tau_1$. This can ... –  Brian M. Scott Feb 6 '13 at 10:38
    
... be done with each pair of the three topologies in your question. (Or four, including the usual Euclidean topology.) –  Brian M. Scott Feb 6 '13 at 10:39
    
@Brain M scott how this logic helps to prove that X is either saperable, first countable and second countable just give me the hints how can I relate these things –  math Feb 10 '13 at 20:56
    
@motu: I gave very big hint for separability: $\Bbb Q^n$ is dense in $\Bbb R^n$ in these topologies. First countability is trivial once you’ve proved that they’re the same topology, because (3) is a metric space, and metric spaces are trivially first countable. Separable metric spaces are second countable, but if you don’t know that theorem, it’s easy to find a countable base for (1), say, and use the fact that the three are the same: there are only countably many boxes that are products of open intervals with rational endpoints. –  Brian M. Scott Feb 10 '13 at 22:37
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On $\mathbb R^n$ the tree are the same.

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@Daien L how could u show that the tree in R^n. Elaborate ur hints –  math Feb 11 '13 at 14:47
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