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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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Do you know what is meant by the term identification topology? –  Brian M. Scott Apr 30 '12 at 21:22
    
I know the definition... –  Manos May 1 '12 at 1:47
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Can you match it up with Matt's sketch? –  Brian M. Scott May 1 '12 at 1:51
    
I can understand Matt's sketch. What i don't see is why these sets form a basis. E.g. if $O$ is an open set of $I/I_o$ containing $a^*$, then its inverse image in $I$ must be open and contain $I_o$. The sets $[0,\epsilon) \cup (1-\epsilon]$ are certainly open and contain $I_o$. But why any pre-image of $O$ can be described as the union of these sets? I must be missing something... –  Manos May 1 '12 at 12:27
    
If $O$ is an open nbhd of $a^*$ in $I/I_o$, the pre-image of $O$ must be an open set $V$ in $I$ containing $I_o$. $0\in V$, and $V$ is open, so $[0,\epsilon_0)\subseteq V$ for some $\epsilon_0>0$. Similarly, $(1-\epsilon_1]\subseteq V$ for some $\epsilon_1>0$. Let $\epsilon=\min\{\epsilon_0,\epsilon_1\}$; then $[0,\epsilon)\cup(1-\epsilon,1]\subseteq V$. Thus, $O$ is an open nbhd of $a^*$ iff its pre-image contains a set of the form $[0,\epsilon)\cup(1-\epsilon,1]$, which is exactly the claim that you quoted. –  Brian M. Scott May 1 '12 at 17:22

2 Answers 2

up vote 7 down vote accepted

Like this:

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That ought to do the trick! –  Brian M. Scott Apr 30 '12 at 21:33
    
@BrianM.Scott ; ) ${}{}{}{}$ –  Matt N. Apr 30 '12 at 21:33

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?

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