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Consider the equivalence relation $‘∼’ $on R × [0, 1] defined by $(x, t) ∼ (x + 1, t), x ∈ R$ and$ t ∈ [0, 1].$ Let $X =(R × [0, 1])/ ∼$ be the quotient space.

Prove that X is Hausdorff and compact.

i was thinking that ,If the quotient map is open, then $X/ \sim$ is a Hausdorff space if and only if $\sim$ is a closed subset of the product space $[0,1]\times X$.

as im very weak in topology and im newly learning topology

pliz help me

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    $\begingroup$ It seems X is homeomorphic to [0,1)×[0,1] which is Hausdorff and not compact. $\endgroup$ – William Elliot Feb 24 '18 at 11:38
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    $\begingroup$ @WilliamElliot $X$ is homeomorphic to $S^1\times[0,1]$. $\endgroup$ – drhab Feb 24 '18 at 11:56
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Let $f:\mathbb R\times[0,1]\to S^1\times[0,1]$ be prescribed by:$$\langle x,t\rangle\mapsto\langle\langle\cos2\pi x,\sin2\pi x\rangle,t\rangle$$

Then $f$ is continuous and surjective.

Further it can be shown that $f$ is a closed map, hence $f$ is a quotient map.

If $q:\mathbb R\times[0,1]\to X$ denotes the quotient map then it can be shown that $f$ and $q$ respect each other in the sense that $f(x,t)=f(y,s)\iff q(x,t)=q(y,s)$.

Based on these facts it can be shown that $X$ and $S^1\times[0,1]$ are homeomorphic, so it remains to prove now that $S^1\times[0,1]$ is compact and Hausdorff.

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  • $\begingroup$ thanks alots .@drhab as i was thinking about how $\mathbb R\times[0,1]homeomorphism S^1\times[0,1]$??? as R is connected but not compact $\endgroup$ – lomber Feb 24 '18 at 12:17
  • $\begingroup$ ok i got its @drhab $\endgroup$ – lomber Feb 24 '18 at 12:26

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