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I want to show that if $(X,\mathcal{T})$ is a metric topology. Take two disjoint subsets $A$ and $B$ such that forall $x\in A$ and $y\in B$ $d(x,y)\geq\delta$. Put the subspace topology on $A,B,A\cup B$. Prove that $A\cup B$ is homeomorphic to $A+B$ (where $(A+B)$ is the disjoint union topology)

My proof bsically involves using the obvious map: $f:A\cup B\rightarrow((A\times \{0\})\cup(B\times\{1\}))$ such that $f(x)=\begin{cases} (x,0) & \mbox{if}\ x\in A \\ (x,1) & \mbox{if} x\in B \end{cases}$

Then I need to apply the inverse of $f$ and $f^{-1}$ to open sets in the topology and show that they are open.

My idea for doing this is to note that as these are metric spaces they have a basis of unions of open balls.

Then if f and $f^{-1}$ map the open balls to open balls then that will be enough to show that they are continuous and so $f$ is a homeomorphism.

Now if I can show that the open balls in $A\cup B$ are all of the form (open balls in A) $\cup$ (open balls in B). That is there is no open ball in $A\cup B$ which is not the union of open balls in $A$ and open balls in $B$.

Then $f$ and $f^{-1}$ obviously map open balls to open balls.

My proof that if $B(a,\epsilon)$ is an open ball in $A\cup B$ then it is the union of an open ball in $A$ and an open ball $B$.:

Take an open ball $B(a,\epsilon)\subset (A\cup B)$ for $a\in A$ or $B$ such that $\exists x,y\in B(a,\epsilon)$ for $x\in A$ and $y\in B$.

Now consider $(B(a,\epsilon)\setminus B)\subset A$. We have to show that this is open (and so it is the union of open balls in $A$). Suppose that it was closed, then:

$\exists x\in (B(a,\epsilon)\setminus B)$ such that $\not\exists\ r>0:B(x,r)\subset A$.

However as $B(a,\epsilon)$ is open in $A\cup B$ then $\exists r>0:B(x,r)\subset (A\cup B)$

Now take $i$ such that $\frac{\delta}{i}<r$ then $B(x,\frac{\delta}{i})\subset A$ and as $B(x,\frac{\delta}{i})\subset B(a,\epsilon)$ it is open. This shows that $B(a,\epsilon)\setminus B$ is open in $A$ and $A\cup B$ which gives the contradiction we require.

We can then use the same argument on $(B(a,\epsilon)\setminus A)$ to show that it is open. So any open ball in $A\cup B$ is a union of open balls in $A$ and open balls in $B$, where the open balls in $A$ and $B$ are open in $A\cup B$.

I this roughly correct (I've only given an outline of most of what I have done)

I have tagged this as homework as this was part of an assessment that I have already handed in and so is not really homework anymore but I'm not really looking for a complete solution to the problem if what I have done is incorrect just a small hint or something

Thanks very much for any help

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You’ve proved that $B(a,\epsilon)$ is the union of an open ball in $A$ and an open set in $B$; this is good enough to get what you want, but it’s not true that $B(a,\epsilon)\cap B$ is necessarily an open ball in $B$.

You’re working harder than necessary, though. Let $$\mathscr{B}=\big\{B(x,\epsilon)\cap(A\cup B):x\in A\cup B\text{ and }0<\epsilon<\delta\big\}\;;$$ then $\mathscr{B}$ is a base for the subspace topology on $A\cup B$. Note that every $V\in\mathscr{B}$ is a subset either of $A$ or of $B$. If $\mathscr{B}_A=\{V\in\mathscr{B}:V\cap B=\varnothing\}$ and $\mathscr{B}_B=\{V\in\mathscr{B}:V\cap A=\varnothing\}$, then $\mathscr{B}=\mathscr{B}_A\sqcup\mathscr{B}_B$, $\mathscr{B}_A$ is a base for $A$, and $\mathscr{B}_B$ is a base for $B$. Now consider

$$\widehat{\mathscr{B}_A}=\big\{V\times\{0\}:V\in\mathscr{B}_A\big\}$$

and

$$\widehat{\mathscr{B}_B}=\big\{V\times\{1\}:V\in\mathscr{B}_B\big\}\;.$$

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But isn't any open set in B a union of open balls in B as there is a basis of open balls of B? –  hmmmm Oct 22 '12 at 14:01
    
So I just need that any open ball in $A\cup B$ is the union of (unions of open balls in A) and (unions of open balls in B). Then when I'm using the map f I can just consider f on open balls? –  hmmmm Oct 22 '12 at 14:02
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@hmmmm: Yes, every open set in $B$ is a union of open balls of $B$. But you claimed that $B(a,\epsilon)$ was the union of an open ball in $A$ and an open ball in $B$, which need not be true. $B(a,\epsilon)\cap B$ is an open set in $B$ but not necessarily an open ball. Yes, you can consider just open balls, or members of any other base for the topology. In my answer I suggested a base that makes the argument about as simple as it can get. –  Brian M. Scott Oct 22 '12 at 14:26
    
Ok, cool but it is the union of (union of open balls in A) and (union of open balls in B)- this is what I should have claimed (not the union of an open ball in A and open ball in B). Thanks for the help –  hmmmm Oct 22 '12 at 14:40
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