Metamath Proof Explorer

Theorem flfssfcf

Description: A limit point of a function is a cluster point of the function. (Contributed by Jeff Hankins, 28-Nov-2009) (Revised by Stefan O'Rear, 9-Aug-2015)

		Ref	Expression
	Assertion	flfssfcf	$⊢ (J \in TopOn (X) \land L \in Fil (Y) \land F : Y ⟶ X) \to (J fLimf L) (F) \subseteq (J fClusf L) (F)$

Proof

Step	Hyp	Ref	Expression
1		flimfcls	$⊢ J fLim (X FilMap F) (L) \subseteq J fClus (X FilMap F) (L)$
2	1	a1i	$⊢ (J \in TopOn (X) \land L \in Fil (Y) \land F : Y ⟶ X) \to J fLim (X FilMap F) (L) \subseteq J fClus (X FilMap F) (L)$
3		flfval	$⊢ (J \in TopOn (X) \land L \in Fil (Y) \land F : Y ⟶ X) \to (J fLimf L) (F) = J fLim (X FilMap F) (L)$
4		fcfval	$⊢ (J \in TopOn (X) \land L \in Fil (Y) \land F : Y ⟶ X) \to (J fClusf L) (F) = J fClus (X FilMap F) (L)$
5	2 3 4	3sstr4d	$⊢ (J \in TopOn (X) \land L \in Fil (Y) \land F : Y ⟶ X) \to (J fLimf L) (F) \subseteq (J fClusf L) (F)$