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Let $H=[-1,1]\times \{0\}$ and $V=\{0\}\times [-1,0)$ in the plane. Let $T=H \cup V$. Show that $T$ is not homeomorphic to the unit interval $I=[0,1]$.

My idea for this problem is that , if we remove a point from the unit interval , we will be left with at most two connected components, but if we remove the origin from $T$ we will be left with $3$ connected components. Is this enough to prove that $I$ and $T$ are not homeomorphic ? How should I write my answer rigirously?

Any help is appreciated, Thanks !

the8thone
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  • Yes, that would be enough. How rigorously do you want to answer? – davidlowryduda Jul 26 '14 at 04:13
  • this is a problem in a previous qualifying exam. I suppose this must be a general theorem that, if removing any point from a topological space $X$ results in at most $m$ connected components and removing a particular point from a topological space $Y$ results in $n$ connected components, then $X$ and $Y$ are not homeomorphic. How can one prove that there is no homeomorphism between these spaces ? – the8thone Jul 26 '14 at 04:19

2 Answers2

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Yes, your proof is correct, but from your comment above, it seems that the reason why it is correct is not completely clear to you. What you are implicitly using is the following proposition.

Proposition: Let $X$ and $Y$ be topological spaces. Suppose there is $x \in X$ such that $X\setminus\{x\}$ has $m$ connected components. If there is no $y \in Y$ such that $Y\setminus\{y\}$ has $m$ connected components, then $X$ and $Y$ are not homeomorphic.

Note: The conditions we need to check are about the spaces $X\setminus\{x\}$ and $Y\setminus\{y\}$, but our conclusion pertains to $X$ and $Y$.

I have included the proof below, you just need to move your mouse over the second grey box to see it, but I recommend you try to prove it yourself first. If you want a hint, move your mouse over the first grey box.

Hint: If $f : X \to Y$ is a homeomorphism, then for any $A \subseteq X$, $f|_A : A \to f(A)$ is a homeomorphism.

Proof: By way of contradiction, suppose that there is a homeomorphism $f : X \to Y$. Then $f|_{X\setminus\{x\}} : X\setminus\{x\} \to Y\setminus\{f(x)\}$ is a homeomorphism. As the number of connected components is a topological invariant, and $X\setminus\{x\}$ has $m$ connected components, $Y\setminus\{f(x)\}$ has $m$ connected components. This is a contradiction as, by assumption, there is no $y \in Y$ such that $Y\setminus\{y\}$ has $m$ connected components.

Michael Albanese
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Fact: If $X$ is homeomorphic to $Y$ with homeomorphism $f: X \longrightarrow Y$, then $X \setminus\{x_0\}$ is homeomorphic to $Y \setminus \{f(x_0)\}$ with the restriction of $f$ giving the homeomorphism.

(In fact, this is direct from the subspace topology.)

If $T$ were homeomorphic to $I$, then $T \setminus\{(0,0)\}$ would be homeomorphic to $I$ minus some point. But it is clear that if we remove the point $\{y_0\}$ from $I$, then the remaining space is covered by connected open sets $[0,y_0)\cup(y_0,1]$ (where maybe one of these is empty if $y_0 = 0$ or $1$). So $I$ minus one point is either connected or has two connected components.

However $T\setminus\{(0,0)\}$ has three connected components, which you can again give explicitly.

Thus $T\setminus\{(0,0)\}$ is not isomorphic to $I$ minus a point, meaning that $T$ is not isomorphic to $I$ (contrapositive of the fact).

davidlowryduda
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