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Let $X$ be a set and $\tau_1,\tau_2$ two topologies on $X$ such that every open set of $\tau_1$ is also open in $\tau_2$. Prove that if $(X,\tau_2)$ is connected then $(X,\tau_1)$ is also connected. Prove this also holds for path connectedness.

My attempt:

Suppose $(X,\tau_1)$ is not connected. Then there are two disjoint nonempty subsets $\theta_1,\theta_2 \in \tau_1$ such that $X=\theta_1 \cup \theta_2$. Every open subset in $\tau_1$ is in $\tau_2$, so we have two disjoint subsets $\theta_1,\theta_2 \in \tau_2$ such that $X=\theta_1 \cup \theta_2$. This contradicts the connectedness of $(X,\tau_2)$.

I'm struggling to prove the result for path connected sets. My definition for path connected sets is:

$(X,\tau)$ is path connected if for any two points $x,y \in X$ there exists a continuous function $f:[0,1] \rightarrow X$ such that $f(0)=x, f(1)=y$.

Path connectedness is not defined in terms of open sets so I don't know how to use the definition in this problem.

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What you did for connectedness is correct.

For path connectedness, don't forget that continuity can be re-defined in terms of open sets. So, if $(X,\tau_2)$ is path connected and if $p,q\in X$, just take $\gamma\colon[0,1]\longrightarrow X$ such that $\gamma(0)=p$, $\gamma(1)=q$, and $\gamma$ is continuous with respect to $\tau_2$. Then $\gamma$ is also continuous with respect to $\tau_1$, because if $A$ is an open subset of $(X,\tau_1)$, then it is also an open subset of $(X,\tau_2)$ and therefore $\gamma^{-1}(A)$ is an open subset of $[0,1]$. So, $\gamma$ is continuous a function from $[0,1]$ into $X$ with respect to $\tau_1$.

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