# prove that the “connected pair” is path-connected

Consider the topological space $$X = \{0, 1\}, τ = \{∅, X, \{0\}\}$$

Show that X is path-connected.

Let $$t ∈]0, 1]$$, I try to prove that the path $$γ : [0, 1] → \{0, 1\}$$ sending $$[0, t[$$ over $$0$$ and $$[t, 1]$$ over 1 is continuous.

But for the open element $$X$$, we have $$γ^{-1}(X)=[0,1]$$ which is not open for the standard topology on $$\mathbb{R}$$?

Thanks for your help.

• You claim that $[0,1]$ is not closed for the standard topology on $\mathbb R$. Sure about that? – José Carlos Santos Jan 24 at 16:05
• I think it's a typo and he meant "open in the standard topology". Note that it should be open in $[0,1]$, not in $\mathbb{R}$. And indeed $[0,1]$ is open in $[0,1]$. – freakish Jan 24 at 16:07
• I see, I was considering $[0,1]\subset \mathbb{R}$. Thank you for your hint! – PerelMan Jan 24 at 16:08
• @JoséCarlosSantos: Thank you, I meant not open – PerelMan Jan 24 at 19:15

## 1 Answer

Just define $$\gamma(t)= 0$$ for all $$t \in [0,1)$$ and $$\gamma(1)=1$$

Then $$\gamma^{-1}[\emptyset]=\emptyset=\emptyset \cap [0,1]$$, $$\gamma^{-1}[\{0\}] = [0,1)=(-1,1) \cap [0,1]$$ and $$\gamma^{-1}[X]=[0,1]=\mathbb{R} \cap [0,1]$$.

All three inverse images of open sets are open in $$[0,1]$$ in the subspace topology as witnessed by the intersection representations.