# Proving a topological space is path connected

Suppose $$(X,T_X)$$ is a topological space and $$\infty_X \notin X$$. Write $$X^* = X \cup \{\infty_X\}$$ and suppose the open sets of $$X^*$$ the empty set and the union of an open set in $$X$$ and the point $$\{\infty_X\}$$.

I want to show that $$(X^*,T_X^*)$$ is a path connected space. This means I have to show that there is a single path connected component ($$X^*$$ itself), i.e. that for any $$x,y \in X^*$$ there is a continuous map $$\gamma: I \rightarrow X$$ where $$\gamma(0) = x, \gamma(1)=y$$. Here is $$I=[0,1]$$ equipped with the Euclidean topology.

I don't really have a clue how to proceed, should I do something with the fact that $$\gamma$$ is continuous and work with the definition? I'm guessing that the special point $$\{\infty_X\}$$ has an important role to show path connectedness but I am confused because we don't have any information on the connectedness of $$X$$.

EDIT: Can we conclude from this that any topological space can be embedded as a closed subspace of a path connected topological space? If so, how would I go about doing that?

Take $$x\in X$$ and define $$\gamma\colon[0,1]\longrightarrow X$$ by$$\gamma(t)=\begin{cases}\infty_X&\text{ if }t\in[0,1)\\x&\text{ otherwise.}\end{cases}$$Then $$\gamma(0)=\infty_X$$ and $$\gamma(1)=x$$. Besides, $$\gamma$$ is continuus since, if $$A$$ is an open subset of $$X$$:

• If $$\infty_X,x\notin A$$, $$\gamma^{-1}(A)=\emptyset$$, which is an open subset of $$[0,1]$$;
• If $$\infty_X\in A$$ and $$x\notin A$$, $$\gamma^{-1}(A)=[0,1)$$, which is an open subset of $$[0,1]$$;
• If $$\infty_X,x\in A$$, $$\gamma^{-1}(A)=[0,1]$$, which is an open subset of $$[0,1]$$.

Since you cannot have $$x\in A$$ and $$\infty_X\notin A$$, this proves that $$\gamma$$ is continuous. And $$\gamma$$ is a path from $$\infty_X$$ to $$x$$. Since there is such a path for every $$x\in X$$, $$X$$ is path-connected.

• What about $x,y \neq \infty_X$? It seems to me you only constructed paths with $\infty_X$... Mar 26 at 20:50
• You take a path from $x$ to $\infty_X$, followed by a path from $\infty_X$ to $y$. Mar 26 at 21:22
• I'm glad I could help. Mar 26 at 21:35
• Or directly, $\gamma(0)=x$, $\gamma(1)=y$, and $\gamma(t)=\infty_X$ otherwise Mar 26 at 21:48
• The continuous injection is the identity map from $X$ into $X^*$ ($x\mapsto x$). Mar 28 at 19:27