The dimension of a topological space $X$ is defined to be the length of the maximal chain of closed irreducible subsets $\varnothing \neq X_0 \subsetneq X_1 \subsetneq ... \subsetneq X_n \subset X$.
Let $A$ be a subset of topological space $X$, how can I show that dim$A$ $\leq$ dim$X$?
I can show that given a chain of closed irreducible subsets of $A$, $A \cap X_0 \subsetneq ... \subsetneq A \cap X_n$, where $X_0,...,X_n$ are closed subsets of X, we can get a chain of closed subsets $X_0 \cap... \cap X_n \subsetneq ... \subsetneq X_{n-1} \cap X_n \subsetneq X_n$ of $X$ of the same length, but I am having trouble showing the subsets are irreducible. Any help is appreciated!