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Let $I=[0,1]$ be the unit interval, viewed as a topological subspace of the real line $R$. Let $I_o=\left\{0,1 \right\}$ be the boundary of $I$. Then denote by $I/{I_o}$ the topological space defined by taking $I$ and shrinking $I_o$ to a point $a^*$ and the topology being the identification topology.

I read the statement "a basis for the open sets of $I/I_o$ containing $a^*$ is the totality of images of the sets of the form $[0,\epsilon) \cup (1-\epsilon,1]$".

How can we see that? Thanks :-)

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    I understand your argument.$I$also think that i misunderstood the meaning of "basis".$I$see now. Thanks a lot!2012-05-01

2 Answers 2

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    @BrianM.Scott ; ) ${}{}{}{}$2012-04-30
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This is not really an answer, not maybe even a hint, but something to ponder about to get a starting motivation for identification topologies.

Consider the function $ f(x)=\sin(2\pi x). $ It is continuous on $I$ and $f(0)=f(1)$ so it certainly defines a function on $I/I_o$. It would be nice if $f(x)$ remained continuous as a function on $I/I_o$, but in order for that to make sense we have to endow $I/I_o$ with a topology.

Such a topology would have the property (by the continuity request for $f(x)$) that the subsets $ A_\epsilon=\{z\in I/I_o\text{ such that }|f(z)|<\epsilon\} $ are open. Can you figure out what these sets $A_\epsilon$ are, concretely?