# topological spaces and some property verification

Which of the following statements are true?

$(a)$ Let $X$ be a set equipped with two topologies $\tau_1$ and $\tau_2$. Assume that any given sequence in $X$ converges with respect to the topology $\tau _1$ if and only if it also converges with respect to the topology $\tau_2$. Then $\tau_1=\tau_2$.

$(b)$ Let $(X,\tau_1)$ and $(Y,\tau_2)$ be two topological spaces and let $f:X\to Y$ be a given map. Then $f$ is continuous if and only if given any sequence $\{x_n\}_{n=1}^\infty$ such that $x_n \rightarrow x$ in $X$, we have $f(x_n) \rightarrow f(x)$ in $Y$.

$(c)$ Let $(X,\tau)$ be a compact topological space and let $\{x_n\}_{n=1}^\infty$ be a sequence in $X$. Then it has a convergent subsequence.

By sequential criteria of continuity, $(b)$ should be true. But the answer given is false, which I don't understand why. Again for $(c)$, Bolzano-Weierstrass theorem states every bounded sequence has a convergent subsequence. Now question is, does every compact set has every sequence bounded in itself? If yes, then the answer would be true. But it is also false as given in the answer. I sense that $(a)$ is not true in general but failed to find a counterexample. What would be the correct logic behind all these properties?

For (b) take the identity from $(X,\tau_1)$ to $(X,\tau_2)$ for the counterexample given in (a).
For (c) note that compact is not equivalent to sequentially compact. For example $[0,1]^{[0,1]}$ with the product topology is compact but not sequentially compact (see What's going on with "compact implies sequentially compact"?)