A graph (G) is only connected as a topological if G is connected as a graph.

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So in our textbook it's stated that "A Graph G is connected as a topological space if and only if G is connected as a graph" without a proof and my lecturer did not prove it during the lecture, so I am wondering how that proof would go.

Thank you for your time!

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Here are some hints:

=> If the graph is connected as a graph, it is path connected as a topological space, hence connected as a topological space.

<= If the graph $G$ is not connected as a graph, separate it into two components (as graphs) $G_1$ and $G_2$. Each of $G_1$ and $G_2$ is a graph. Now here's a fact:

Every graph can be embedded in $\mathbb R^3$.

But since $\mathbb R^3$ is homeomorphic to any of its open balls, every graph can be embedded in any open ball of $\mathbb R^3$. Now take two disjoint open balls $B_1$ and $B_2$ of $\mathbb R^3$, and embed each $G_i$ in $B_i$.

Another way to do it if you're familiar with CW complexes: a graph as a topological space is a $1$-dimensional CW complex. We know that CW complexes are locally path connected. But any connected space that is locally path connected is path connected.