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?
Best Answer
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 $X$:
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.